SC-6 Elementary Companion Problem Workbook

COMPANION PROBLEM VOLUME // FROM BLOCKS TO THE STACK (SC-6) ● 8 CHAPTERS • 253 WORKED REAL-WORLD PROBLEMS • RPNCORE VERIFIED

Companion Practice & Problem-Solving Volume to From Blocks to the Stack: Mastering Elementary Mathematics with Physical Manipulatives and the SC-6 RPN Calculator**

This companion workbook houses the 253 extended chapter-by-chapter worked problems, fluency drills, and applied STEM lab challenges that accompany the Bookstore Edition of From Blocks to the Stack. Every problem below pairs a hands-on physical Gridfinity manipulative setup with an engine-verified SC-6 keystroke sequence and 4-register $(T, Z, Y, X)$ stack trace.

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Chapter Quick-Jump & Manipulative Index

Chapter	Topic & Companion Problem Set	Primary Manipulative Kit	Worked Problems	Textbook Reference
Chapter 02	How Stack Math Works: Stacking Numbers Just Like Blocks	`PLACE-01` Place Value Studio	21	Manual Ch. 02 →
Chapter 03	Counting, Number Lines & Place Value Studio	`PLACE-01` Place Value Studio	24	Manual Ch. 03 →
Chapter 04	Whole Number Operations & Skip-Counting Ladders	`PLACE-01` & `PRIME-01`	30	Manual Ch. 04 →
Chapter 05	Fractions, Decimals, Percents & Proportional Towers	`FRAC-01` Fraction Towers	38	Manual Ch. 05 →
Chapter 06	Geometry, Surface Area & Modulo-60 Time	`FRAC-01` & `PRIME-01`	33	Manual Ch. 06 →
Chapter 07	Discrete Probability, Primes & Pre-Algebra	`PRIME-01` & `FRAC-01`	52	Manual Ch. 07 →
Chapter 08	The Math Quiz Engine & Two-Method Fluency	`PLACE-01`, `FRAC-01` & `PRIME-01`	29	Manual Ch. 08 →
Chapter 09	Applied Elementary STEM & Maker Labs	Full Manipulative Bench	26	Manual Ch. 09 →

Chapter 02: How Stack Math Works: Stacking Numbers Just Like Blocks — Extended Worked Problems

Accompanies Chapter 02 of From Blocks to the Stack (21 Worked Practice Problems)

Workbook Problem 2.1: The Three-Day Gravel Mountain

To fill the fossil-dig sandbox, your crew wheels loads of washed pea gravel from the supply pile: 4 wheelbarrows on Monday, 5 on Tuesday, and 6 on Wednesday. How many total wheelbarrows of gravel did you pour in? Stack 4 unit cubes on your desk, add 5 cubes (making 9), and then add 6 more cubes to form a tower of 15 cubes.

Watch how the stack chains all three days without clearing: 4 [ENTER] 5 [+] 6 [+] $\to$ 15 (T: 0, Z: 0, Y: 0, X: 15). Why didn't you need to press [ENTER] before typing 6? Because automatic stack lift pushed Tuesday's running total ($9$) upstairs to Floor $Y$ the instant your finger tapped 6!

Workbook Problem 2.2: Stepping Stones Across the Ten-Bridge

Your crew is laying round sensory stepping stones along a winding garden path: 8 smooth slate stones in the morning, and 7 sparkling granite stones in the afternoon. How many stepping stones lead to the climbing tower? Place [8] in Row 1 and [7] in Row 0 of your Place Value Tray (PLACE-01). Take 2 from 7 to complete a full 10 (snapping a Gold [+1] carry token into the Tens column), leaving [5] in the Ones column ($10 + 5 = 15$).

Fuse the morning and afternoon stone deliveries on the stack: 8 [ENTER] 7 [+] $\to$ 15 (T: 0, Z: 0, Y: 0, X: 15), bridging cleanly past ten to reach 15 stepping stones.

Workbook Problem 2.3: Climbing-Dome Steel Bolts

A hardware box holds 10 heavy steel bolts for the geodesic climbing dome. Your crew bolts 6 of them into the main arch. How many bolts are left in the box for the balance-beam brackets? On your Place Value Tray (PLACE-01), start with 10 ([1] in Tens, [0] in Ones) and subtract [6]—finding the complement [4] in the Ones column.

Verify the ten-partner on your stack: 10 [ENTER] 6 [-] $\to$ 4 (T: 0, Z: 0, Y: 0, X: 4), leaving 4 bolts ready in Register $X$.

Workbook Problem 2.4: The Zero-Flag Mystery (Additive Identity)

Your crew stakes 7 bright orange flags along the zip-line landing zone. The safety inspector checks the tape measure and says zero extra flags are needed ($+0$). How many flags stand along the runway? Place 7 counting tiles on your desk and add nothing to them—you still have 7 tiles!

Test the Zero Addition Rule on the stack: 7 [ENTER] 0 [+] $\to$ 7 (T: 0, Z: 0, Y: 0, X: 7), proving that adding $0$ leaves any number unchanged ($a + 0 = a$).

Workbook Problem 2.5: Zipline Carabiners on the Rack

The gear chest has 8 climbing carabiners clipped to the rack. Because the zip-line cable is still being tensioned, zero carabiners are removed this morning ($-0$). How many carabiners stay clipped to the rack? Set out 8 counting tiles and take away zero—all 8 tiles stay right where they are.

Confirm on the keypad: 8 [ENTER] 0 [-] $\to$ 8 (T: 0, Z: 0, Y: 0, X: 8), showing that subtracting zero leaves your starting number intact ($a - 0 = a$).

Workbook Problem 2.6: Clearing the Tire-Swing Chains

Six heavy link chains are uncoiled on the grass for the three playground tire swings. Your crew hoists and bolts all 6 chains onto the overhead beam. How many unbolted chains are left on the grass? Place 6 counting tiles on the desk, then sweep all 6 away into your hand—zero remain ($6 - 6 = 0$).

Run the self-subtraction on your calculator: 6 [ENTER] 6 [-] $\to$ 0 (T: 0, Z: 0, Y: 0, X: 0). Whenever a number subtracts itself, the ground shelf drops cleanly to zero!

Workbook Problem 2.7: The Twin-Pod Explorer Doubler

A glass sky-elevator carries 16 young space-camp explorers up from the lobby. At the sky-bridge level, a twin pod docks and an equal team of 16 more explorers steps aboard! How many explorers are riding together now? Why type 16 twice when [ENTER] clones whatever is on Floor $X$? Watch how 16 [ENTER] [+] doubles a number in two clicks: 16 [ENTER] [+] $\to$ 32 (T: 0, Z: 0, Y: 0, X: 32). Pressing [ENTER] copies 16 onto Floor $Y$ while keeping 16 on Floor $X$, so [+] doubles it to 32 without re-keying!

Workbook Problem 2.8: Starlight Observation Deck Countdown

The glass-floored Starlight Observation Deck holds at most 100 visitors at a time. If a group of 43 stargazers steps out onto the glass floor, how many more visitors can join them before the deck is full? Picture a 100-flat grid: taking away 4 full rows of ten ($40$) leaves $60$, and taking away 3 more units leaves $57$ ($43 + 57 = 100$).

Verify the open spots on the stack: 100 [ENTER] 43 [-] $\to$ 57 (T: 0, Z: 0, Y: 0, X: 57), confirming 57 open stargazer spots.

Workbook Problem 2.9: Express 10-Floor Beacon Test

The tower flight engineer tests the altitude beacons every 10 floors starting from the underground launch bay (Floor 0) up to Floor 50. Line up 5 ten-rods end-to-end on your desk ($10, 20, 30, 40, 50$). Seed the initial 10-floor step on your SC-6: 10 [ENTER] 0 [+] $\rightarrow$ 10 (and in 2-LV mode—or after pressing [g] [REP] on 10 in 4-stack mode—four more taps of [+] climb straight through Floors 20, 30, 40, and 50).

Keying 10 [ENTER] 0 [+] sets the first 10-floor beacon at (T: 0, Z: 0, Y: 10, X: 10) in 2-LV mode so each tap of [+] steps up another 10 stories!

Workbook Problem 2.10: Hover-Drone Charging Bays

The Sky-Tower's drone hangar has 4 launch decks, and each deck has 6 magnetic wireless charging pads. How many drone charging pads are in the hangar altogether? Snap together all 24 counting tiles from your studio kit into a $4 \times 6$ rectangle—your fingers can feel that 4 rows of 6 make a solid rectangle of 24!

Multiply the 4 decks by 6 pads per deck: 4 [ENTER] 6 [×] $\to$ 24 (T: 0, Z: 0, Y: 0, X: 24), giving 24 charging bays.

Workbook Problem 2.11: Robotic Window-Climbing Gecko Bots

To polish the solar-glass windows on the outside of the tower, robotic suction-cup "Gecko Bots" clean 7 giant window panels on each of 8 upper floors. How many window panels do the robots polish on this run? Build a small $3 \times 4$ tile array ($12$) and spin it $90^\circ$ to see why $7 \times 8$ and $8 \times 7$ cover the exact same area. Then multiply 7 panels across 8 floors: 7 [ENTER] 8 [×] $\to$ 56 (T: 0, Z: 0, Y: 0, X: 56), totaling 56 gleaming solar windows!

Workbook Problem 2.12: Sky-Lab Weather Sensor Deployment

The chief meteorologist unpacked 56 wind-speed sensors and mounted an equal number on each of 7 sky-bridge levels. How many sensors guard each sky-bridge? Just as $7 \times 8 = 56$ in Workbook Problem 2.11, dividing $56$ into $7$ equal groups reveals the missing partner side of the rectangle ($56 \div 7 = 8$): 56 [ENTER] 7 [÷] $\to$ 8 (T: 0, Z: 0, Y: 0, X: 8), placing 8 wind sensors on each sky-bridge.

Workbook Problem 2.13: Maker Fair Backpack Swaps

You walk into the Maker Fair with 14 3D-printed badges in your backpack. You win 5 more badges at the robotics booth, then trade 3 badges to a friend. How many badges are in your backpack now? Place 14 counting tiles on your desk, slide 5 more in ($19$), and slide 3 away ($16$).

Track the live running balance on the stack: 14 [ENTER] 5 [+] 3 [-] $\to$ 16 (T: 0, Z: 0, Y: 0, X: 16). Because the subtotal $19$ automatically lifts to Floor $Y$ when you type $3$, you never press [ENTER] a second time!

Workbook Problem 2.14: Battle-Bot Arena Bumper Railings

The Maker Fair's rectangular Battle-Bot arena measures 12 feet long and 8 feet wide. How many feet of polycarbonate bumper railing are needed to wrap all the way around the arena? Why add $12 + 8 + 12 + 8$ four times when a rectangle has two equal L-shaped half-perimeters? Add one length and one width ($12 + 8 = 20\text{ ft}$), then double it: 12 [ENTER] 8 [+] 2 [×] $\to$ 40 (T: 0, Z: 0, Y: 0, X: 40), giving 40 feet of arena bumper railing!

Workbook Problem 2.15: Sailboat Pennant Triangle Area

A triangular racing pennant for a model RC sailboat has a base of 10 cm and a height of 6 cm. What is the area of the pennant? Form a $4 \times 3$ rectangle of 12 tiles and slice it diagonally in half ($12 \div 2 = 6$). Every triangle is exactly half of its surrounding rectangle! Multiply base by height and divide by 2 ($10 \times 6 \div 2 = 30\text{ cm}^2$): 10 [ENTER] 6 [×] 2 [÷] $\to$ 30 (T: 0, Z: 0, Y: 0, X: 30), giving a $30\text{ cm}^2$ pennant area.

Workbook Problem 2.16: The $\$50$ Market Pouch Countdown

You bring a crisp $\$50$ bill to the weekend market. You spend $\$18$ on a crystal-growing kit and $\$12$ on fresh berries. How much cash is left in your pouch? Start with 5 ten-rods ($50$), remove 18 (leaving 32), and remove 12 (leaving 20). (Mental Math Shortcut: Notice that $\$18 + \$12 = \$30$ spent in all, and $\$50 - \$30 = \$20$!)

Deduct both purchases in a continuous chain on the stack: 50 [ENTER] 18 [-] 12 [-] $\to$ 20 (T: 0, Z: 0, Y: 0, X: 20), leaving $\$20.00$ in your pouch.

Workbook Problem 2.17: Splitting the Giant Nacho Platter

Four friends split a giant stadium nacho platter costing $\$17.50$. Using Quotient & Remainder ([g] [÷R]), find how many whole one-dollar bills each friend pays and how much coin change is left to split! Deal out 17 dollar tiles among 4 friends: each friend puts in 4 whole dollars ($\$16.00$), leaving $\$1.50$ to cover with quarters!

Split $\$17.50$ across 4 friends with [g] [÷R]: 17.50 [ENTER] 4 [g] [÷R] $\to$ 4 (T: 0, Z: 0, Y: 1.5, X: 4), showing $X = \$4.00$ whole dollars each and $Y = \$1.50$ remaining coin change.

Workbook Problem 2.18: Rescuing a Number from the Top Floor

Suppose you parked the number 10 way up on the top shelf ($T = 10$) while 20, 30, and 40 sit below it. How many clicks of [f] [R↓] bring the top shelf ($T$) down to the active ground shelf ($X$)? Three clicks ([f] [R↓] [f] [R↓] [f] [R↓]) rotate $T \to Z \to Y \to X$: 10 [ENTER] 20 [ENTER] 30 [ENTER] 40 [f] [R↓] [f] [R↓] [f] [R↓] $\rightarrow$ 10.

Loading all four shelves and rotating three times lands at (T: 20, Z: 30, Y: 40, X: 10), bringing 10 cleanly into Register $X$.

Workbook Problem 2.19: Crushing the Giant Fraction Bar

Look at this intimidating fraction bar: $$\frac{48 - (4 \times 6)}{3 + (2 \times 4.5)}$$ On a normal calculator, you would need four sets of parentheses! On your SC-6, work from the inside out: compute the top ($48 - 24 = 24$), compute the bottom ($3 + 9 = 12$) while $24$ waits upstairs, and press [÷]: 48 [ENTER] 4 [ENTER] 6 [×] [-] 2 [ENTER] 4.5 [×] 3 [+] [÷] $\to$ 2 (T: 0, Z: 0, Y: 0, X: 2), resolving the entire fraction bar to $2.0$ with zero parentheses!

Workbook Problem 2.20: The Treehouse Zip-Line Shortcut (Pythagorean Theorem)

Your backyard treehouse sits $6\text{ meters}$ north and $8\text{ meters}$ east of the porch post, forming a right triangle with legs of 6 and 8. How long is the diagonal zip-line cable $c = \sqrt{6^2 + 8^2}$ connecting them? Square both legs ([x²]), add them together ([+]), and tap Square Root ([f] [√x]): 6 [x²] 8 [x²] [+] [f] [√x] $\to$ 10 (T: 0, Z: 0, Y: 0, X: 10), finding a $10.0\text{ m}$ zip-line in 7 keystrokes with zero scratch paper!

Workbook Problem 2.21: Weather Station Thermometer Code ($^\circ\text{F} \to ^\circ\text{C}$)

Your school weather station reads $68^\circ\text{F}$. Convert it to Celsius using the inside-out formula $C = (F - 32) \times 5 \div 9$: 68 [ENTER] 32 [-] 5 [×] 9 [÷] $\to$ 20 (T: 0, Z: 0, Y: 0, X: 20), landing cleanly on $20.0^\circ\text{C}$.

Chapter 03: Counting, Number Lines & Place Value Studio — Extended Worked Problems

Accompanies Chapter 03 of From Blocks to the Stack (24 Worked Practice Problems)

Workbook Problem 3.1: Bioluminescent Jellyfish Pods

A marine tank holds 8 neon-blue comb jellies in Pod A and 7 glowing moon jellies in Pod B. How many bioluminescent jellyfish are swimming in the tank? On your Place Value Tray (PLACE-01), [8] needs 2 more to complete 10—telling you right away to split 7 into $2 + 5$ and snap a Gold [+1] carry token into the Tens column!

Bridge to ten on the stack: 8 [ENTER] 2 [+] 5 [+] $\to$ 15 (T: 0, Z: 0, Y: 0, X: 15), filling the first ten-group with 2 moon jellies before placing the remaining 5 into Frame 2's top row. $X = 15$ (1 full ten-group plus 1 full five-row).

Workbook Problem 3.2: Completing the Double-Frame Solar Array

Your nature dome's rooftop charger has 16 mini solar tiles wired across two 10-slot mounting frames. How many more solar tiles do you need to snap in so both ten-groups are completely full at 20? On your Place Value Tray (PLACE-01), 16 has [1] in the Tens column and [6] in the Ones column—needing 4 more ones to regroup to 20.

Snap the 4 remaining solar tiles into the second tray: 16 [ENTER] 4 [+] $\to$ 20 (T: 0, Z: 0, Y: 0, X: 20). Both frames are now 100% full at $X = 20$ because $6 + 4$ are ten-group partners!

Workbook Problem 3.3: Combining Lunar Sample Canisters

Inside the clean-room airlock, Rover Tray A holds 9 sealed lunar dust canisters and Rover Tray B holds 8 canisters. How many canisters are ready for analysis? Because 9 is just 1 away from 10, borrow 1 canister from the 8 (splitting 8 into $1 + 7$) to complete the first ten-group!

Total the two canister trays: 9 [ENTER] 1 [+] 7 [+] $\to$ 17 (T: 0, Z: 0, Y: 0, X: 17), completing Frame 1 at 10 and leaving 7 in Frame 2 ($X = 17$).

Workbook Problem 3.4: Five-Anchor Addition on the Bottom Wire

Your underwater docking bay already holds 12 waterproof camera drones (10 on Rail 1, and 2 red beads on Rail 2). A surface boat lowers 5 more camera drones down the cable. How many drones are in the bay now? On your Place Value Tray (PLACE-01), place 12 ([1] in Tens, [2] in Ones) in Row 1 and [5] in Row 0 of the Ones column to combine the ones ($2 + 5 = 7$, giving 17)!

Add the 5 new drones to the bay tally: 12 [ENTER] 5 [+] $\to$ 17 (T: 0, Z: 0, Y: 0, X: 17), combining the two groups into 17 without ever touching the top rail of 10 ($X = 17$).

Workbook Problem 3.5: Decomposing Teen Numbers on the Place Value Tray

A reef diver carries a mesh bag with 18 biodegradable coral-reef temperature tags (10 on Rail 1, and 8 on Rail 2: 5 red + 3 white). Along the outer reef wall, the diver clips 5 tags onto brain corals. How many tags are left in the mesh bag? On Rail 2, slide a whole group of 5 beads back to the right in one motion, leaving 3 beads on Rail 2 and all 10 on Rail 1!

Subtract the 5 attached reef tags: 18 [ENTER] 5 [-] $\to$ 13 (T: 0, Z: 0, Y: 0, X: 13), leaving 10 on Wire 1 and 3 on Wire 2 ($X = 13$).

Workbook Problem 3.6: Splitting a Teen Number into Ten and Ones

A sea-floor basket holds 15 titanium diving weights. To store 10 weights in the main ballast locker and keep the rest on the workbench, how many single weights stay on the bench? On the Place Value Tray (PLACE-01), 15 is [1] in the Tens socket and [5] in the Ones socket.

Subtract the full bundle of 10 to isolate the ones: 15 [ENTER] 10 [-] $\to$ 5 (T: 0, Z: 0, Y: 0, X: 5), proving that taking 10 away from 15 leaves the 5 red beads on Wire 2!

Workbook Problem 3.7: Clearing the Bottom Rail in One Sweep

A deep-sea sample rack holds 19 glowing hydrothermal vent tubes (10 on Rail 1, and 9 on Rail 2). Scientists transfer 8 of those tubes into the lab freezer. How many vent tubes remain in the sample rack? On Rail 2, sliding 8 beads back to the right leaves just 1 lonely red bead on the bottom wire alongside the full 10 on the top wire.

Subtract the 8 transferred tubes: 19 [ENTER] 8 [-] $\to$ 11 (T: 0, Z: 0, Y: 0, X: 11), leaving 10 on Wire 1 and 1 on Wire 2 ($X = 11$).

Workbook Problem 3.8: Subtracting from a Full Twenty-Rack

A scuba gear locker holds a full rack of 20 compressed-air tanks across two rows of 10. A dive team checks out 7 tanks for an afternoon kelp-forest survey. How many tanks remain in the locker? On your Place Value Tray (PLACE-01), start with 20 ([2] Tens, [0] Ones) and regroup 1 ten into 10 ones so you can subtract 7 ($10 - 7 = 3$, leaving 13).

Find the remaining scuba tanks on the rack: 20 [ENTER] 7 [-] $\to$ 13 (T: 0, Z: 0, Y: 0, X: 13). Even without counting, you know $10 - 7 = 3$ on the bottom wire, plus the untouched 10 on the top wire gives $X = 13$!

Workbook Problem 3.9: Measuring the Gap Between Desert Landmarks

On your trail map, the Ghost Gum tree stands at kilometer marker 15, and Red Sand Gap stands at kilometer marker 42. How far apart are these two landmarks? Remember that subtraction doesn't just mean "taking away"—on a number line, it measures the distance gap between two points! Mentally hop from 15 to 20 ($+5$), from 20 to 40 ($+20$), and from 40 to 42 ($+2$).

Calculate the exact gap on the stack: 42 [ENTER] 15 [-] $\to$ 27 (T: 0, Z: 0, Y: 0, X: 27), confirming the $5 + 20 + 2 = 27\text{ km}$ trail interval ($X = 27$).

Workbook Problem 3.10: Solar Rover Odometer Leap

Your solar-powered desert rover drives 35 kilometers across a dry salt flat in the morning, then rolls another 18 kilometers through red dunes in the afternoon. What does the odometer read at camp? On your Open Number Line, start at 35, use 5 of the 18 kilometers to reach round milestone 40, and leap the remaining 13 kilometers to 53!

Log the rover's full-day distance: 35 [ENTER] 18 [+] $\to$ 53 (T: 0, Z: 0, Y: 0, X: 53), bridging across decades 40 and 50 to land at $X = 53\text{ km}$.

Workbook Problem 3.11: Camp Water Tank Countdown

Your desert camp starts with a 50-liter solar water tank. Over two hot days on the trail, the crew drinks and cooks with 24 liters of water. How many liters remain in the tank? On your Open Number Line, start at 50, leap back 20 units to hit 30, and step back 4 more units to land on 26.

Subtract the 24 liters used on the trail: 50 [ENTER] 24 [-] $\to$ 26 (T: 0, Z: 0, Y: 0, X: 26), leaving $X = 26\text{ liters}$ in the tank.

Workbook Problem 3.12: Double-Zone Outback Tree Survey

A wildlife botanist counts 28 river red gums along the northern creek bed and 36 ghost gums across the southern gorge. How many native trees are mapped in all? On your Open Number Line, start at 28, hop $+2$ to land on round decade 30, and then leap the remaining $+34$ in one clean bound!

Combine the two tree surveys: 28 [ENTER] 36 [+] $\to$ 64 (T: 0, Z: 0, Y: 0, X: 64), landing squarely on $X = 64$ native trees.

Workbook Problem 3.13: Solar Viewer Nickel Tokens Ladder

At the solar telescope viewing deck, student astronomers count 5-cent tokens up to 50 cents. In 2-Stack mode, each tap of [+] drops another 5-cent group onto the stack: 5 [ENTER] 5 [+] $\rightarrow$ 10.

In 2-Stack mode, run this sequence $\to$ (T: 0, Z: 0, Y: 5, X: 10). Successive taps of [+] step by 5s on the Place Value Tray (PLACE-01), advancing through 15, 20, 25, up to 50 cents (10 nickels = 50 cents).

Workbook Problem 3.14: Planetarium Dime-Rod Ladder

To buy a new lunar filter, the astronomy club stacks 10-cent dimes up to 100 cents (one full dollar). In 2-Stack mode, each press of [+] adds another ten-rod until 10 rods snap together into a 100-flat: 10 [ENTER] 10 [+] $\to$ 20.

In 2-Stack mode, run this sequence $\to$ (T: 0, Z: 0, Y: 10, X: 20). Successive taps of [+] advance by 10s until 10 rods assemble into a complete 100-flat ($X = 100$).

Workbook Problem 3.15: Seven-Day Star-Watch Schedule with `[g] [REP]`

Observatory sky-surveys are planned in 7-day weekly cycles. In 4-Stack mode, fill the stack with 7 using [g] [REP] and step through 7 weeks of star-watch nights by tapping 7 [g] [REP] [+] [+].

In 4-Stack mode, tap 7 [g] [REP] [+] $\to$ (T: 7, Z: 7, Y: 7, X: 14). Five more taps of [+] pull constant 7 from the top of the stack, climbing through 21, 28, 35, 42, to 49 days (7 weeks = 49 days).

Workbook Problem 3.16: The Mystery of the Silent Zero Placeholders

A deep-space radio dish records a burst of 8 thousands and 4 tens of signal pulses—with 0 in the HUNDREDS socket and 0 in the ONES socket of your Place Value Studio (PV-01) Whole Numbers Tray. When you add $8,000$ and $40$ on the stack, why doesn't the number squash together into $84$?

Add the thousands and tens directly: 8000 [ENTER] 40 [+] $\to$ 8040 (T: 0, Z: 0, Y: 0, X: 8040). The two 0 digit miniblocks act like placeholders holding the HUNDREDS and ONES sockets open so the 8 stays in the THOUSANDS column and the 4 stays in the TENS column!

Results: Thousands ($1000$): $8$, Hundreds ($100$): $0$, Tens ($10$): $4$, Ones ($1$): $0$.

Workbook Problem 3.17: Zooming Inside a Single Millimeter

Place value doesn't stop at the ONES socket! Switch to your Decimal Place Value Tray (PV-01: 1 • 0.1 | 0.01 | 0.001): moving right past the raised decimal point (•), each column divides by 10 into tenths ($0.1$), hundredths ($0.01$), and thousandths ($0.001$). A laser micrometer measures a telescope mirror thickness of $345$ whole millimeters, plus $6$ tenths ($0.6$) and $7$ hundredths ($0.07$).

Sum the whole millimeters, tenths, and hundredths: 345 [ENTER] 0.6 [+] 0.07 [+] $\to$ 345.67 (T: 0, Z: 0, Y: 0, X: 345.67), assembling whole units, tenths, and hundredths into $345.67\text{ mm}$.

Tenths place $= 6$, Hundredths place $= 7$.

Workbook Problem 3.18: Dimes and Pennies as Decimal Blocks

In the observatory gift shop, a glow-in-the-dark Saturn sticker costs 4 dimes ($0.40$) and 7 pennies ($0.07$). Because a dime is $\frac{1}{10}$ of a dollar and a penny is $\frac{1}{100}$ of a dollar, pocket change is a real-life decimal place-value kit!

Add the dimes and pennies together: 0.40 [ENTER] 0.07 [+] $\to$ 0.47 (T: 0, Z: 0, Y: 0, X: 0.47), combining 4 dimes and 7 pennies into $\$0.47$.

Results: Dimes (Tenths) $= 4$, Pennies (Hundredths) $= 7$.

Workbook Problem 3.19: Combining Planetarium Seating Bowls

The science center's outdoor star-amphitheater seats $10,000$ sky-watchers, and the indoor laser-dome seats $2,500$ more. Combine these two large place-value blocks on the stack to find the total seating capacity.

Total the outdoor and indoor seating capacities: 10000 [ENTER] 2500 [+] $\to$ 12500 (T: 0, Z: 0, Y: 0, X: 12500), combining both venues into $12,500$ seats.

Results: Ten-thousands $= 1$, Thousands $= 2$, Hundreds $= 5$.

Workbook Problem 3.20: Eclipse Watchers Across Three Cities

During a total solar eclipse, three neighboring towns log $400,000$, $50,000$, and $200$ sky-watchers wearing eclipse glasses. Combine the three place-value tiers on the stack and notice which columns stay at zero!

Combine the three sky-watcher tiers: 400000 [ENTER] 50000 [+] 200 [+] $\to$ 450200 (T: 0, Z: 0, Y: 0, X: 450200), building $450,200$ with zero thousands, zero tens, and zero ones.

Results: Hundred-thousands $= 4$, Ten-thousands $= 5$, Hundreds $= 2$.

Workbook Problem 3.21: Human Heartbeats in a Full Decade

At a steady resting pulse of 80 beats per minute, your heart beats about $42,048,000$ times in a single year! What happens to every digit's place-value column when you multiply by $10\text{ years}$ to find your heartbeats across a full decade?

Multiply the annual heartbeat count by 10 years: 42048000 [ENTER] 10 [×] $\to$ 420480000 (T: 0, Z: 0, Y: 0, X: 420480000), shifting every digit one column to the left to reach $420,480,000$ beats!

Decompose Hundred-Millions: [ENTER] 100000000 [PV] $\longrightarrow X = \mathbf{4}$ (4 hundred-million heartbeats).

Workbook Problem 3.22: Counting Water Drops in an Olympic Pool

An Olympic-sized swimming pool holds $2,500,000\text{ liters}$ of water, and every single liter holds about $20,000\text{ water drops}$. Multiply those two giant numbers on the stack to discover how many drops fill the pool, then spotlight the ten-billions digit!

Multiply the liters by drops per liter: 2500000 [ENTER] 20000 [×] $\to$ 50000000000 (T: 0, Z: 0, Y: 0, X: 50000000000), revealing an astonishing $50,000,000,000$ ($50\text{ billion}$) water drops!

Decompose Ten-Billions: [ENTER] 10000000000 [PV] $\longrightarrow X = \mathbf{5}$ (5 ten-billion drops).

Workbook Problem 3.23: Spotlight on Saturn's Giant Rings

Through the observatory's main telescope, Saturn's glittering ring system spans $282,000\text{ kilometers}$ from edge to edge—wide enough to line up more than 22 Earths side by side! Use [PV] to spotlight the hundred-thousands, ten-thousands, and thousands digits: 282000 [ENTER] 100000 [PV] 10000 [PV] 1000 [PV] $\rightarrow$ 2.

Place Value Spotlight: Park 282,000 in Register $Y$ with 282000 [ENTER]. Because [PV] never erases Register $Y$, you can inspect each column in a chain: 100000 [PV] yields 2 hundred-thousands, 10000 [PV] yields 8 ten-thousands, and 1000 [PV] yields 2 thousands (Y: 282000, X: 2).

Results: Hundred-Thousands $= 2$, Ten-Thousands $= 8$, Thousands $= 2$ (a palindrome pattern: $2\text{-}8\text{-}2$!).

A city library network holds $875,420$ adventure and science books shared equally across 12 branch libraries. First, spotlight the ten-thousands digit (875420 [ENTER] 10000 [PV] $\longrightarrow X = \mathbf{7}$). Then calculate the average number of books per branch: 875420 [ENTER] 12 [÷] $\rightarrow$ 72,951.66667.

This sequence yields an average branch collection of $\mathbf{72,951.67\text{ books}}$ (T: 0, Z: 0, Y: 0, X: 72951.66667).

Chapter 04: Whole Number Operations & Skip-Counting Ladders — Extended Worked Problems

Accompanies Chapter 04 of From Blocks to the Stack (30 Worked Practice Problems)

Workbook Problem 4.1: Remapping the Twenty-Four Tiles into Four Rows

Can those exact same 24 tiles form a chunkier rectangle without any leftovers? Rearrange your Cobalt Blue and White tiles into 4 rows with 6 tiles in each row to model a 4-by-6 arcade button grid.

Find the total button count on the stack: 4 [ENTER] 6 [×] $\rightarrow$ 24. This sequence lifts the 4 rows into Register $Y$, multiplies by 6 buttons per row, and displays $(T: 0, Z: 0, Y: 0, X: 24)$ buttons ($X = 24$). Pause and notice: $3 \times 8$ and $4 \times 6$ look like completely different rectangles, yet both hold the exact same 24 tiles!

Workbook Problem 4.2: The Five-Column Pixel Sprite Matrix

In your game studio, you design a glowing treasure-map icon made of 7 rows with 5 Radiant Gold pixel tiles in each row. Because every row is a friendly Five-Anchor, you can count by fives ($5, 10, 15, 20, 25, 30, 35$) or multiply the rectangle in one strike!

Multiply the 7 rows by 5 pixels per row: 7 [ENTER] 5 [×] $\rightarrow$ 35. This sequence holds the 7 rows in Register $Y$, multiplies by 5 pixels per row, and evaluates directly to $(T: 0, Z: 0, Y: 0, X: 35)$ pixels ($X = 35$).

Workbook Problem 4.3: Six-Row Geode Display Case

A rock-and-mineral collector builds a wooden wall case with 6 shelves and 9 glowing crystal compartments on each shelf. On your desk mat, build a smaller $2 \times 9 = 18$-tile starter section first, then imagine stacking three of those $2 \times 9$ sections together!

Scale the calculation to all 6 shelves on the stack: 6 [ENTER] 9 [×] $\rightarrow$ 54. This sequence loads 6 shelves into Register $Y$ and multiplies by 9 compartments, yielding $(T: 0, Z: 0, Y: 0, X: 54)$ crystals ($X = 54$).

Workbook Problem 4.4: The Classic Eight-by-Eight Chessboard Square

Why is $64$ called a square number? Because when a rectangle has the exact same number of rows and columns ($8$ rows of $8$ tiles), it forms a geometric square! Build a $4 \times 4 = 16$-tile checkerboard corner using alternating white and slate tiles on your desk mat, then scale up to a full $8 \times 8$ chessboard on the SC-6.

Compute the full $8 \times 8$ board area: 8 [ENTER] 8 [×] $\rightarrow$ 64. This sequence lifts 8 into Register $Y$ and multiplies by 8 to drop the stack to $(T: 0, Z: 0, Y: 0, X: 64)$ squares ($X = 64$). (Shortcut tip: Since both factors are 8, you can also just type 8 [ENTER] [×] to square 8 in three taps!)

Workbook Problem 4.5: Hydroponic Space-Garden Tower

Inside a lunar habitat greenhouse, a vertical hydroponic rack holds 12 columns with 4 lettuce pods in each column. Snap together 3 columns of 4 tiles ($12$ tiles) on your desk mat—then notice that 12 columns is just four of those $3 \times 4$ blocks side by side!

Find the pod capacity across all 12 columns: 12 [ENTER] 4 [×] $\rightarrow$ 48. This sequence parks the 12 columns in Register $Y$ and multiplies by 4 pods per column, leaving $(T: 0, Z: 0, Y: 0, X: 48)$ plants thriving in the lunar greenhouse ($X = 48$).

Workbook Problem 4.6: Granite Cliff Anchor Ropes by Fours

To reach a ledge high on a granite cliff, a rescue climber clips an anchor bolt every 4 vertical meters. Build 4 four-tile rungs ($16$ tiles) on your desk mat and slide your finger up the rungs: 4, 8, 12, 16 meters.

Find the total rope height at the 9th anchor bolt: 4 [g] [REP] [+] [+] [+] [+] [+] [+] [+] [+] $\rightarrow$ 36. Pressing 4 [g] [REP] loads the 4-meter step across the entire stack, and 8 rhythmic taps of [+] climb to $(T: 4, Z: 4, Y: 4, X: 36)$ meters ($X = 36\text{ meters}$).

Workbook Problem 4.7: High-Altitude Shelter Packs by Sixes

Eight mountain huts are stocked with 6 emergency solar blankets each. On your Place Value Tray (PLACE-01), place [6] in Row 1 and [6] in Row 0 ($6 + 6 = 12$, regrouping a Gold [+1] carry token to Tens) to see the first two huts.

Step through all 8 mountain huts with Constant Repeat: 6 [g] [REP] [+] [+] [+] [+] [+] [+] [+] $\rightarrow$ 48. Seeding the stack with 6 [g] [REP] keeps 6 locked in Register $T$, while 7 steady taps of [+] climb to $(T: 6, Z: 6, Y: 6, X: 48)$ solar blankets ($X = 48$).

Workbook Problem 4.8: Glacier Rope-Bridge Carabiners by Eights

To cross a crevasse field, the rescue team rigs 8 rope-bridge spans, using 8 locking steel carabiners on every span. Assemble 2 rows of 8 counting tiles ($16$ tiles) on your desk mat to see the first two spans.

Climb the 8s ladder across all 8 spans: 8 [g] [REP] [+] [+] [+] [+] [+] [+] [+] $\rightarrow$ 64. Replicating 8 with 8 [g] [REP] and tapping [+] 7 times marches through the 8s table ($16, 24, 32, 40, 48, 56$) to $(T: 8, Z: 8, Y: 8, X: 64)$ carabiners ($X = 64$).

Workbook Problem 4.9: Summit Radio Beacons and the Magic Nine-Pattern

Nine mountain summits each broadcast 9 emergency weather pulses per minute. On your Place Value Tray (PLACE-01), compare adding [9] to adding [1] Ten and stepping the Ones digit back by 1 ($9 + 9 = 18$)!

Climb all 9 summits with Constant Repeat: 9 [g] [REP] [+] [+] [+] [+] [+] [+] [+] [+] $\rightarrow$ 81. Priming all four registers with 9 [g] [REP] and tapping [+] 8 times reaches $(T: 9, Z: 9, Y: 9, X: 81)$ pulses ($X = 81$). Watch the screen as you climb ($09, 18, 27, 36, 45, 54, 63, 72, 81$): the tens digit climbs up by 1 while the ones digit steps down by 1, so the two digits always add up to 9!

Workbook Problem 4.10: Helicopter Dozens Ladder to One Gross

A rescue helicopter drops a supply crate holding 12 boxes of hand-warmers, with 12 warmers in every box (a dozen dozens, called 1 gross!). Snap together a 12-tile strip on your desk and add a second 12-tile strip to reach 24.

Step through all twelve dozens on the stack: key in 12 [g] [REP] to seed all four registers $(T: 12, Z: 12, Y: 12, X: 12)$, followed by 11 rhythmic taps of [+]. The stack advances through every dozen ($24, 36, 48 \dots 132$) until reaching 144 warming packets ($X = 144$).

Workbook Problem 4.11: Five-Stage Compound Pulley Doubler

A cliff-rescue winch starts with a $3\text{ kN}$ pulling force. Every time you clip on another compound pulley block, the pulling force doubles! Start with 3 tiles on your desk and double them to 6, 12, and 24 tiles.

Double 3 through 5 pulley stages ([ENTER] [+]): 3 [ENTER] [+] [ENTER] [+] [ENTER] [+] [ENTER] [+] [ENTER] [+] $\rightarrow$ 96. Tapping [ENTER] [+] five times rockets through $6 \to 12 \to 24 \to 48 \to 96$, leaving $(T: 0, Z: 0, Y: 0, X: 96)\text{ kN}$ on the screen ($X = 96\text{ kN}$)—32 times stronger than the bare rope!

Workbook Problem 4.12: Trail Mix Energy Bars in Eight-Packs

Your expedition kitchen bakes 75 oat-and-honey trail bars and seals them into waterproof pouches holding 8 bars each. How many full 8-packs can you seal, and how many bars are left over for the cooks to snack on?

Pack all 75 bars into pouches of 8: 75 [ENTER] 8 [g] [÷R] $\rightarrow$ 9. Integer division loads quotient 9 into Register $X$ and remainder 3 into Register $Y$ with inline trace $(T: 0, Z: 0, Y: 3, X: 9)$ ($Q = 9\text{ pouches}$, $R = 3\text{ leftover bars}$).

Workbook Problem 4.13: The Four-Person Wilderness Tent Dilemma

Twenty-nine student explorers hike up to base camp for an overnight star-watch, and each expedition tent sleeps 4 campers. When you divide 29 by 4, why isn't the answer just the quotient?

Find the tent count on the stack: 29 [ENTER] 4 [g] [÷R] $\rightarrow$ 7. The stack drops into $(T: 0, Z: 0, Y: 1, X: 7)$. Seven tents hold 28 campers, leaving $Y = 1$ camper still standing outside! Because you can't leave a friend out in the cold, any remainder $> 0$ means you must pitch $7 + 1 = \mathbf{8\text{ tents}}$ ($Q = 7\text{ full tents}$, $R = 1\text{ camper}$, total $= 8\text{ tents}$).

Workbook Problem 4.14: Cutting Parachute Cord from a Master Spool

To rig model recovery parachutes, you cut $35\text{ cm}$ shroud lines from a $250\text{ cm}$ spool of braided nylon cord. How many full $35\text{ cm}$ lines can you cut, and how many centimeters of cord are left on the spool?

Divide the $250\text{ cm}$ spool into $35\text{ cm}$ lines: 250 [ENTER] 35 [g] [÷R] $\rightarrow$ 7. Tapping [g] [÷R] yields $(T: 0, Z: 0, Y: 5, X: 7)$, giving 7 full parachute lines with a $5\text{ cm}$ scrap left over ($Q = 7\text{ lines}$, $R = 5\text{ cm leftover}$).

Workbook Problem 4.15: Nine-Grid Meteorite Display Trays

The planetarium curator sorts 115 iron meteorite fragments into $3 \times 3$ display boxes that hold 9 fragments each. How many complete 9-fragment boxes can be filled?

Package all 115 meteorites into 9-slot boxes: 115 [ENTER] 9 [g] [÷R] $\rightarrow$ 12. The display resolves into $(T: 0, Z: 0, Y: 7, X: 12)$, filling 12 complete display boxes with 7 fragments left over for the touch-table ($Q = 12\text{ boxes}$, $R = 7\text{ fragments}$).

Workbook Problem 4.16: Whitewater River Rafting Boats

Forty-six adventure campers arrive at the river dock, and each inflatable whitewater raft holds 7 paddlers. How many rafts get completely filled, and how many rafts must launch so everyone gets down the river?

Divide all 46 campers across the 7-passenger rafts: 46 [ENTER] 7 [g] [÷R] $\rightarrow$ 6. Integer division drops the registers to $(T: 0, Z: 0, Y: 4, X: 6)$; since the 4 remaining campers in Register $Y$ need a boat too, the river guides launch $6 + 1 = \mathbf{7\text{ rafts}}$ ($Q = 6\text{ full rafts}$, $R = 4\text{ campers}$, total $= 7\text{ rafts}$).

Workbook Problem 4.17: Building Four-Wheel All-Terrain Rovers

The robotics workshop bin holds 78 knobby rubber tires for 4-wheeled mini-rovers. Unlike tents or rafts (where leftover people need an extra vehicle), can 2 leftover tires build a whole 4-wheel rover on their own?

Sort all 78 tires into 4-wheel sets: 78 [ENTER] 4 [g] [÷R] $\rightarrow$ 19. The stack drops into $(T: 0, Z: 0, Y: 2, X: 19)$, proving you can build 19 complete 4-wheel rovers with 2 spare tires left in the parts bin ($Q = 19\text{ rovers}$, $R = 2\text{ spare tires}$).

Workbook Problem 4.18: Weather-Station Four-Battery Packs

Each automated forest weather station runs on a 4-battery pack, and your field kit has 42 fresh rechargeable AA batteries. How many stations can you power up completely?

Check how many 4-battery packs 42 batteries can fill: 42 [ENTER] 4 [g] [÷R] $\rightarrow$ 10. Integer division leaves $(T: 0, Z: 0, Y: 2, X: 10)$ on the screen: 10 weather stations get full 4-battery packs, leaving 2 spare batteries in your backpack ($Q = 10\text{ packs}$, $R = 2\text{ batteries}$).

Workbook Problem 4.19: Eight-Pack Glow-Stick Bundles for Cave Explorers

Before entering a limestone cavern, the expedition leader bundles 175 chem-light glow-sticks into team packs of 8 sticks each. How many full 8-packs are ready, and how many extra glow-sticks go to the lead guide?

Divide all 175 glow-sticks into packs of 8: 175 [ENTER] 8 [g] [÷R] $\rightarrow$ 21. The operational stack separates $175 \div 8$ into $(T: 0, Z: 0, Y: 7, X: 21)$, making 21 full team packs with 7 extra sticks ($Q = 21\text{ packs}$, $R = 7\text{ glow-sticks}$).

Workbook Problem 4.20: Claw-Bot Servos & Bumper Sensors

Your grabbing claw-bot needs 5 high-torque digital servos at $\$12$ each and 4 bumper touch-sensors at $\$8$ each. How much does the whole sensor-and-servo upgrade cost?

Combine the servo and sensor subtotals on the stack: 5 [ENTER] 12 [×] 4 [ENTER] 8 [×] [+] $\rightarrow$ 92. Multiplying $5 \times 12$ stores $\$60$ while $4 \times 8 = \$32$ evaluates in Register $X$; tapping [+] fuses both subtotals into $(T: 0, Z: 0, Y: 0, X: 92)$, giving $\mathbf{\$92}$ ($X = 92$).

Workbook Problem 4.21: 3D-Printed Mini-Drone Frame Fundraiser

For the school maker fair, your club 3D-prints 45 mini-drone frames that sell for $\$2$ each, and you spend $\$28$ on spools of neon PLA filament. What is your club's net profit after paying for the filament?

Find the net fundraiser profit on the stack: 45 [ENTER] 2 [×] 28 [-] $\rightarrow$ 62. Evaluating $45 \times 2 = \$90$ and immediately subtracting the $\$28$ filament cost leaves $(T: 0, Z: 0, Y: 0, X: 62)$, earning $\mathbf{\$62}$ for the club ($X = 62$).

Workbook Problem 4.22: Electric Go-Kart Frame Rails & Corner Brackets

To build a solar go-kart frame, your team orders 40 feet of aluminum frame tubing at $\$6$ per foot, plus 5 heavy-duty corner-bracket packs at $\$14$ per pack. Chain both products and add them on the stack!

Calculate the combined go-kart frame order: 40 [ENTER] 6 [×] 5 [ENTER] 14 [×] [+] $\rightarrow$ 310. Multiplying $40 \times 6 = \$240$ lifts automatically to Register $Z$ while you compute $5 \times 14 = \$70$, and the final [+] drops the stack into $(T: 0, Z: 0, Y: 0, X: 310)$ ($X = \$310$).

Workbook Problem 4.23: The Robot Pit-Crew Bolt Bin Tracker

Your team's pit-crew organizer bin starts tournament morning with 850 hex bolts. During round-one builds, 145 bolts are used on new bumpers, and then 88 spare bolts from a retired practice bot are recycled back into the bin. How many hex bolts are in the bin now?

Track the bolt bin on the stack: 850 [ENTER] 145 [-] 88 [+] $\rightarrow$ 793. Subtracting 145 leaves 705 bolts, and adding the 88 recycled bolts leaves $(T: 0, Z: 0, Y: 0, X: 793)$ bolts ready for the finals ($X = 793$).

Workbook Problem 4.24: Three-Part Arcade Cabinet Wiring Kit

To wire a custom classroom arcade cabinet, you buy three batches of parts: 120 LED jumper wires at $\$3$ per pack, 85 arcade micro-switches at $\$2$ each, and 40 braided USB cables at $\$4$ each. Watch the stack accumulate all three products in a chain!

Sum all three arcade wiring batches on the stack: 120 [ENTER] 3 [×] 85 [ENTER] 2 [×] [+] 40 [ENTER] 4 [×] [+] $\rightarrow$ 690. The stack evaluates jumper wires ($120 \times 3 = \$360$), adds switches ($85 \times 2 = \$170 \to \$530$), and adds cables ($40 \times 4 = \$160$) to finish at $(T: 0, Z: 0, Y: 0, X: 690)$ ($X = \$690$).

Workbook Problem 4.25: Packaging Brass Standoffs for Robot Brains

Four maker tables each build 24 microcontroller boards, and every board needs 2 brass standoff posts. If standoffs come in bags of 15, how many full bags do you empty, and how many extra standoffs do you need from one more bag?

Chain the multiplication and remainder division on the stack: 4 [ENTER] 24 [×] 2 [×] 15 [g] [÷R] $\rightarrow$ 12. Chaining $4 \times 24 \times 2$ produces 192 standoffs, and tapping 15 [g] [÷R] drops the stack into $(T: 0, Z: 0, Y: 12, X: 12)$: 12 full bags are used, plus 12 standoffs from a 13th bag ($Q = 12\text{ bags}$, $R = 12\text{ standoffs}$, open 13 bags).

Workbook Problem 4.26: The Ninety-Nine Overshoot Shortcut

When a tournament timer flashes $45 \times 99$, don't do long multiplication! Ninety-nine groups of 45 is just 100 groups of 45 minus 1 group of 45 ($45 \times 100 - 45$).

Run the $100\times$ minus $1\times$ shortcut on the stack: 45 [ENTER] 100 [×] 45 [-] $\rightarrow$ 4455. Keying 45 [ENTER] 100 [×] gives $4,500$, and subtracting one group of 45 [-] immediately lands on $(T: 0, Z: 0, Y: 0, X: 4455)$!

Workbook Problem 4.27: Young Gauss's Staircase Pairing Trick

What is the sum of every whole number from $1$ to $20$ ($1 + 2 + 3 + \dots + 20$)? Build a 4-step staircase with 10 counting tiles ($1 + 2 + 3 + 4 = 10$), build a second identical 10-tile staircase, flip it upside down, and snap the two staircases together! They form a solid $4 \times 5 = 20$ rectangle—and half of that rectangle is $10$! For $1$ to $20$, two staircases snap into a $20 \times 21$ rectangle, and half of that area is the sum!

Evaluate Gauss's pairing rectangle for $1$ through $20$: 20 [ENTER] 21 [×] 2 [÷] $\rightarrow$ 210. Entering 20 [ENTER] 21 [×] builds the $420$ double-staircase rectangle, and pressing 2 [÷] cuts it in half to $(T: 0, Z: 0, Y: 0, X: 210)$!

Workbook Problem 4.28: The Secret of Squaring Numbers Ending in Five

Here is a mental-math party trick that looks like wizardry: to square any two-digit number ending in 5 (like $35^2$), take the tens digit ($3$), multiply it by the next whole number up ($3 \times 4 = 12$), and stick $25$ on the end: $1,225$! Why does it work? Because $(30 + 5) \times (30 + 5) = 30 \times 30 + 2 \times (30 \times 5) + 25 = 30 \times 40 + 25 = 1,200 + 25$!

Verify $35^2$ on the stack: 35 [ENTER] 35 [×] $\rightarrow$ 1225. Keying 35 [ENTER] 35 [×] (or simply 35 [ENTER] [×] without re-typing 35!) yields $(T: 0, Z: 0, Y: 0, X: 1225)$. Now try $65^2$ in your head ($6 \times 7 = 42 \to 4,225$) and check it with 65 [ENTER] [×]!

Workbook Problem 4.29: Why Odd Numbers Always Build Perfect Squares

What happens when you add the first 10 odd numbers ($1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19$)? Try it with your counting tiles first: place 1 tile in the corner. Wrap an L-shaped border of 3 tiles around it—you get a $2 \times 2 = 4$ square! Wrap 5 more tiles around the outside—you get a $3 \times 3 = 9$ square! Wrap 7 more tiles—you get a $4 \times 4 = 16$ square! Every new odd number adds one new row and one new column, so the sum of the first $10$ odd numbers is simply a $10 \times 10$ square!

Square $10$ on the stack to find the 10-odd-number sum: 10 [ENTER] 10 [×] $\rightarrow$ 100. Keying 10 [ENTER] 10 [×] (or 10 [ENTER] [×]) yields $(T: 0, Z: 0, Y: 0, X: 100)$!

Workbook Problem 4.30: The Zero-Remainder Divisibility Detector

How can you tell in one second whether $16$ is an exact factor of $4,896$? Use [g] [÷R] as a divisibility detector: if Register $Y$ (the remainder) lands on 0, the division is $100\%$ exact!

Test whether 4,896 divides evenly by 16: 4896 [ENTER] 16 [g] [÷R] $\rightarrow$ 306. Keying 4896 [ENTER] 16 [g] [÷R] sets Register $X$ to 306 and Register $Y$ to 0 (T: 0, Z: 0, Y: 0, X: 306). Remainder 0 proves that $4,896 = 16 \times 306$ with zero leftovers!

Chapter 05: Fractions, Decimals, Percents & Proportional Towers — Extended Worked Problems

Accompanies Chapter 05 of From Blocks to the Stack (38 Worked Practice Problems)

Workbook Problem 5.1: Simplifying the Championship Pizza Party Tally

At the end-of-season pizza celebration, the kitchen serves giant 18-slice sheet pizzas, and your table devours $\frac{48}{18}$ pizzas in total! Model this with your Fraction Studio kit by placing four $1/6$ blocks ($56.0\text{ mm}$) beside two $1/3$ blocks ($56.0\text{ mm}$) to see how grouping slices shrinks the numbers without changing the amount of pizza, then reduce $\frac{48}{18}$ and unpack the whole pies on your SC-6: [.] 48 [.] 18 [ENTER] [g] [SIMP] $\rightarrow$ 8/3.

Stack the slice count . 48 . 18 $(T: 0, Z: 0, Y: 0, X: 48/18)$ and [ENTER] $(T: 0, Z: 0, Y: 48/18, X: 48/18)$, factor out $6$-slice clusters with [g] [SIMP] $(T: 0, Z: 0, Y: 48/18, X: 8/3)$, and unpack the whole pizzas via [a ᵇ/c] $(T: 0, Z: 0, Y: 48/18, X: 2\ 2/3)$—showing $2\ \frac{2}{3}$ full sheet pizzas!

Workbook Problem 5.2: Tuning the Rocket Propellant Digital Gauge

Your model rocket's water-pressure booster tank is filled to $\frac{5}{8}$ of its capacity, but the launchpad's digital pressure computer asks for a decimal readout. Stack five $1/8$ towers in the comparator well ($52.5\text{ mm}$) and turn the blocks to their right-hand decimal face to spot how each eighth adds $0.125$, then flip the fraction into a decimal on the stack: [.] 5 [.] 8 [ENTER] [FDISP] $\rightarrow$ 0.625.

Lock in the tank level with . 5 . 8 and [ENTER] $(T: 0, Z: 0, Y: 5/8, X: 5/8)$, then flip the display format with [FDISP] $(T: 0, Z: 0, Y: 5/8, X: 0.625)$ to see why five eighths ($5 \times 0.125$) equals $0.625$ on the launch computer!

Workbook Problem 5.3: Reading the Telemetry Sensor Back into Eighths

After liftoff, a propellant flow sensor reports $0.375$ liters remaining in the reserve bottle. Align a $31.5\text{ mm}$ height with three cyan $1/8$ blocks in your comparator well to see why $3 \times 0.125 = 0.375$, then translate the sensor's decimal straight back into a clean fraction for your flight log: [.] 375 [ENTER] [FDISP] $\rightarrow$ 3/8.

Put the telemetry reading .375 onto the stack with [ENTER] $(T: 0, Z: 0, Y: 0.375, X: 0.375)$ and tap [FDISP] $(T: 0, Z: 0, Y: 0.375, X: 3/8)$ to watch $375$ thousandths snap cleanly into $3/8$!

Workbook Problem 5.4: Stop-Motion Animation Reel Timing

You are filming a claymation dinosaur scene that lasts $2\ \frac{3}{4}$ seconds, and your video timeline editor measures clips in decimal seconds. Stack two Whole towers plus three cobalt-blue $1/4$ blocks ($231\text{ mm} = 2.75 \times 84\text{ mm}$) to see the three quarters sitting at the $0.75$ mark, then toggle your reel length into decimal seconds: [2] [.] 3 [.] 4 [ENTER] [FDISP] $\rightarrow$ 2.75.

Place the mixed scene time 2 . 3 . 4 on the stack with [ENTER] $(T: 0, Z: 0, Y: 2\ 3/4, X: 2\ 3/4)$ and tap [FDISP] $(T: 0, Z: 0, Y: 2\ 3/4, X: 2.75)$ to match the video timeline's $2.75$-second marker!

Workbook Problem 5.5: The Endless Echo of One-Third on the Metronome

A musical triplet splits a single beat into 3 equal third-notes ($\frac{1}{3}$ of a beat each). Drop one solar-yellow $1/3$ tower ($28.0\text{ mm}$) into the comparator well: because $10$ and $100$ are only divisible by $2$ and $5$, no stack of tenths or hundredths will ever line up flush with $\frac{1}{3}$! Check how the SC-6 displays this infinite decimal march: [.] 1 [.] 3 [ENTER] [FDISP] $\rightarrow$ 0.3333333333.

Load one triplet beat . 1 . 3 with [ENTER] $(T: 0, Z: 0, Y: 1/3, X: 1/3)$ and switch views via [FDISP] $(T: 0, Z: 0, Y: 1/3, X: 0.3333333)$ to light up the daylight screen's repeating bar over the digit $3$ ($0.\bar{3}$).

Workbook Problem 5.6: Shrink-Raying the Skateboard Ramp Score

In the neighborhood skateboard jam, a rider lands $35$ out of $50$ trick attempts ($\frac{35}{50}$). Line up 7 tenth-bars ($7 \times 8.4\text{ mm} = 58.8\text{ mm}$) in the comparator channel to see how grouping by $5$s turns fiftieths into tenths, then use hardware [g] [SIMP] on your SC-6 to peel away the common factor: [.] 35 [.] 50 [ENTER] [g] [SIMP] $\rightarrow$ 7/10.

Record the trick tally . 35 . 50 with [ENTER] $(T: 0, Z: 0, Y: 35/50, X: 35/50)$ and fire [g] [SIMP] $(T: 0, Z: 0, Y: 35/50, X: 7/10)$ to divide top and bottom by $5$, revealing a clean $7/10$ landing rate!

Workbook Problem 5.7: Counting Full Pizzas from Twenty-Three Slices

Your tabletop board-game club orders medium pizzas cut into $6$ slices each, and the hungry players eat $\frac{23}{6}$ pizzas. Stack six emerald $1/6$ blocks ($6 \times 14.0\text{ mm} = 84.0\text{ mm}$) next to 1 Whole tower to see how every $6$ slices build a full pie (and why $23$ slices is just $1$ slice shy of $4$ full pies!), then unpack $\frac{23}{6}$ into a mixed number: [.] 23 [.] 6 [ENTER] [a ᵇ/c] $\rightarrow$ 3 5/6.

Stack the $23$ sixth-slices . 23 . 6 with [ENTER] $(T: 0, Z: 0, Y: 23/6, X: 23/6)$ and tap [a ᵇ/c] $(T: 0, Z: 0, Y: 23/6, X: 3\ 5/6)$ to separate the $3$ whole pizzas from the $5/6$ remaining pie!

Workbook Problem 5.8: Finding the Master Rhythm Block with Hardware `[GCF]`

An electronic drum machine loops a $24$-step melody against a $36$-step bassline, creating the fraction $\frac{24}{36}$. Using Prime Studio blocks, the longest single rod that measures both $24$ and $36$ with zero overhang is the $12$-unit rod. Use hardware [GCF] on your SC-6 to reveal that greatest common factor in one tap: 24 [ENTER] 36 [GCF] $\rightarrow$ 12.

This extracts $\gcd(24, 36) = 12$ $(T: 0, Z: 0, Y: 0, X: 12)$, showing why dividing both pattern lengths by $12$ simplifies $24/36$ straight to $\mathbf{2/3}$.

Workbook Problem 5.9: Fitting the Twin Lookout Window Trim Pieces

Cutting window trim for the treehouse lookout openings requires two pieces of trim, each $\frac{2}{4}\text{ foot}$ long. Place two $2/4$ block pairs side-by-side with the 84 mm Whole tower in the Comparator Well to feel their flush alignment, then add them on your SC-6 to watch the calculator preserve the unreduced fourths before simplifying: [.] 2 [.] 4 [ENTER] [.] 2 [.] 4 [+] $\rightarrow$ 4/4.

Place the first trim piece . 2 . 4 on the stack with [ENTER] $(T: 0, Z: 0, Y: 2/4, X: 2/4)$, join its twin . 2 . 4 [+] $(T: 0, Z: 0, Y: 0, X: 4/4)$ so you can verify all four fourths ($\frac{4}{4}$), and tap [g] [SIMP] $(T: 0, Z: 0, Y: 0, X: 1)$ to fuse them into $1\text{ whole foot}$.

Workbook Problem 5.10: Blending Pine Resin and Linseed Roof Sealant

Mixing weatherproofing roof sealant in your workshop bucket requires combining $\frac{3}{4}\text{ gallon}$ of natural pine resin with $\frac{2}{5}\text{ gallon}$ of boiled linseed oil. In the Comparator Well, a $3/4$ tower ($63\text{ mm}$) and a $2/5$ tower ($33.6\text{ mm}$) reach $96.6\text{ mm}$—rising past the 84 mm Whole tower! Let the SC-6 bridge fourths and fifths into twentieths automatically: [.] 3 [.] 4 [ENTER] [.] 2 [.] 5 [+] [a ᵇ/c] $\rightarrow$ 1 3/20.

Pour in the resin . 3 . 4 with [ENTER] $(T: 0, Z: 0, Y: 3/4, X: 3/4)$, blend the linseed oil . 2 . 5 [+] $(T: 0, Z: 0, Y: 0, X: 23/20)$ to see $\frac{15}{20} + \frac{8}{20} = \frac{23}{20}$, and tap [a ᵇ/c] $(T: 0, Z: 0, Y: 0, X: 1\ 3/20)$ to read $1\ \frac{3}{20}\text{ gallons}$.

Workbook Problem 5.11: Lining Up Ridge-Beam and Ceiling Bolt Holes

You need to drill pilot holes for structural rafter bolts spaced every $\frac{1}{8}\text{ foot}$ along the ridge beam and every $\frac{1}{12}\text{ foot}$ along the ceiling brace. Laying $8$-unit and $12$-unit rods side-by-side shows they first meet flush at $24$ units (rather than $8 \times 12 = 96$). Press hardware [f] [LCM] to find that exact common denominator: [8] [ENTER] 12 [f] [LCM] $\rightarrow$ 24.

This confirms that eighths and twelfths meet at 24ths, with $\text{LCD} = \mathbf{24}$ $(T: 0, Z: 0, Y: 0, X: 24)$.

Workbook Problem 5.12: Raising the Lookout Balcony Safety Rail

Measuring support timbers for the lookout balcony railing involves joining a $1\ \frac{1}{2}\text{-foot}$ base post with a $2\ \frac{1}{3}\text{-foot}$ upper safety rail extension. Stack 1 Whole block plus a $1/2$ block ($42\text{ mm}$) alongside 2 Whole blocks plus a $1/3$ block ($28\text{ mm}$) in the wells: the fractional tops combine to $70\text{ mm}$, matching five emerald $1/6$ blocks ($5 \times 14\text{ mm} = 70\text{ mm}$)! Verify on the stack: [1] [.] 1 [.] 2 [ENTER] [2] [.] 1 [.] 3 [+] [a ᵇ/c] $\rightarrow$ 3 5/6.

Set the base post 1 . 1 . 2 into $Y$ with [ENTER] $(T: 0, Z: 0, Y: 1\ 1/2, X: 1\ 1/2)$, bolt on the upper rail 2 . 1 . 3 [+] $(T: 0, Z: 0, Y: 0, X: 23/6)$, and unpack the whole feet via [a ᵇ/c] $(T: 0, Z: 0, Y: 0, X: 3\ 5/6)$ to reach $3\ \frac{5}{6}\text{ feet}$.

Workbook Problem 5.13: Notching the Pine Fascia Strip Around the Trunk

You have a $\frac{3}{4}\text{-foot}$ pine fascia strip and trim off $\frac{1}{4}\text{ foot}$ to clear the tree's main trunk. In the Comparator Well, lifting one $1/4$ block off a stack of three $1/4$ blocks leaves two $1/4$ blocks—which stands flush with a single $1/2$ block! Watch the SC-6 show the unreduced $\frac{2}{4}$ before you simplify: [.] 3 [.] 4 [ENTER] [.] 1 [.] 4 [-] $\rightarrow$ 2/4.

Hold the starting strip . 3 . 4 in $Y$ with [ENTER] $(T: 0, Z: 0, Y: 3/4, X: 3/4)$, saw off the trunk notch . 1 . 4 [-] $(T: 0, Z: 0, Y: 0, X: 2/4)$, and tap [g] [SIMP] $(T: 0, Z: 0, Y: 0, X: 1/2)$ to reveal $1/2\text{ foot}$.

Workbook Problem 5.14: Sawing the Diagonal Cross-Brace from a Heavy Stud

From a $5\ \frac{1}{3}\text{-foot}$ heavy framing stud, you saw off $2\ \frac{3}{4}\text{ feet}$ to build a diagonal cross-brace. Notice why this problem trips up paper-and-pencil calculators: $\frac{1}{3}$ ($4/12$) is smaller than $\frac{3}{4}$ ($9/12$), so you have to trade one whole foot into $12$ twelfths ($16/12 - 9/12 = 7/12$)! Watch the SC-6 handle that regrouping automatically: [5] [.] 1 [.] 3 [ENTER] [2] [.] 3 [.] 4 [-] [a ᵇ/c] $\rightarrow$ 2 7/12.

Clamp the full stud 5 . 1 . 3 into $Y$ with [ENTER] $(T: 0, Z: 0, Y: 5\ 1/3, X: 5\ 1/3)$, saw off the cross-brace 2 . 3 . 4 [-] $(T: 0, Z: 0, Y: 0, X: 31/12)$, and tap [a ᵇ/c] $(T: 0, Z: 0, Y: 0, X: 2\ 7/12)$ to read the $2\ \frac{7}{12}\text{-foot}$ remnant.

Workbook Problem 5.15: Sealing the Trapdoor Sill with Canvas Weather-Stripping

You have 1 full roll of heavy canvas weather-stripping for the treehouse trapdoor, and you cut off $\frac{3}{8}$ of the roll to seal the upper sill. Place one 84 mm Whole tower in Well A and three $1/8$ blocks ($31.5\text{ mm}$) in Well B: because $1\text{ Whole} = \frac{8}{8}$, the gap above Well B fits exactly five $1/8$ blocks ($52.5\text{ mm}$): [1] [ENTER] [.] 3 [.] 8 [-] $\rightarrow$ 5/8.

Start with the full roll 1 [ENTER] $(T: 0, Z: 0, Y: 1, X: 1)$ and slice away the sill strip . 3 . 8 [-] $(T: 0, Z: 0, Y: 0, X: 5/8)$ to leave $5/8$ of the roll.

Workbook Problem 5.16: Zeno's Rope Ladder & the Halfway-to-the-Trapdoor Paradox

The ancient Greek philosopher Zeno of Elea once posed a famous riddle: if you climb halfway to the treehouse trapdoor ($\frac{1}{2}$ the height), then climb half of the remaining gap ($\frac{1}{4}$), and then half of that remaining gap ($\frac{1}{8}$), how close to the top have you reached—and why is the gap left over always equal to the very last step you took? Drop a $1/2$ block ($42\text{ mm}$), a $1/4$ block ($21\text{ mm}$), and a $1/8$ block ($10.5\text{ mm}$) into the Comparator Well: each new block cuts the empty space to the top of the $84\text{ mm}$ Whole tower in half, reaching $73.5\text{ mm}$ with exactly one $1/8$ block ($10.5\text{ mm}$) of space remaining! Chain Zeno's three steps on the stack: [.] 1 [.] 2 [ENTER] [.] 1 [.] 4 [+] [.] 1 [.] 8 [+] $\rightarrow$ 7/8.

Climb the first half . 1 . 2 with [ENTER] $(T: 0, Z: 0, Y: 1/2, X: 1/2)$, halve the remaining gap with . 1 . 4 [+] $(T: 0, Z: 0, Y: 0, X: 3/4)$, and halve it once more with . 1 . 8 [+] $(T: 0, Z: 0, Y: 0, X: 7/8)$ to stand at $7/8$ of the climb—just $\frac{1}{8}$ shy of the trapdoor!

Workbook Problem 5.17: Accounting for the Saw Blade's Bite on Bike Rack Slats

Building oak wall racks to hang customer bikes requires cutting mounting slats measuring $18\ \frac{3}{8}\text{ in}$ each, and the spinning table-saw blade turns an extra $\frac{1}{8}\text{ in}$ of wood into sawdust (the kerf) on every cut! In the Comparator Well, three $1/8$ blocks plus one $1/8$ block fuse into four $1/8$ blocks—a clean $\frac{1}{2}\text{ inch}$: 18 [.] 3 [.] 8 [ENTER] [.] 1 [.] 8 [+] [a ᵇ/c] $\rightarrow$ 18 1/2.

Place the slat length 18 . 3 . 8 into $Y$ with [ENTER] $(T: 0, Z: 0, Y: 18\ 3/8, X: 18\ 3/8)$, add the saw kerf . 1 . 8 [+] $(T: 0, Z: 0, Y: 0, X: 148/8)$, and toggle to mixed form via [a ᵇ/c] $(T: 0, Z: 0, Y: 0, X: 18\ 1/2)$ so you know every slat consumes $18\ \frac{1}{2}\text{ inches}$ of board.

Workbook Problem 5.18: How Many Slats Fit in an Eight-Foot Oak Board?

Now that each slat plus its saw cut takes $18\ \frac{1}{2}\text{ inches}$, how many full slats can you cut from an $8$-foot ($96\text{-inch}$) oak board? Five slats take a bit over $90\text{ inches}$, while six would pass $108\text{ inches}$. Divide the board by the per-slat allowance to check: 96 [ENTER] 18 [.] 1 [.] 2 [÷].

Load the full board 96 [ENTER] $(T: 0, Z: 0, Y: 96, X: 96)$ and divide by the slat-plus-kerf length 18 . 1 . 2 [÷] $(T: 0, Z: 0, Y: 0, X: 5.189)$, proving you can cut $5$ complete slats with a short tail piece left over.

Workbook Problem 5.19: Measuring the Leftover Oak Tail Piece

How long is that leftover oak tail piece after sawing $5$ full slats ($5 \times 18\ \frac{1}{2}\text{ in}$) from the $96\text{-inch}$ board? Compute the wood consumed first, then use [x⇄y] to subtract it from $96\text{ inches}$ without writing down intermediate numbers: [5] [ENTER] 18 [.] 1 [.] 2 [×] 96 [x⇄y] [-] [a ᵇ/c] $\rightarrow$ 3 1/2.

Multiply the $5$ slats 5 [ENTER] $(T: 0, Z: 0, Y: 5, X: 5)$ by 18 . 1 . 2 [×] $(T: 0, Z: 0, Y: 0, X: 92\ 1/2)$, then bring in the $96\text{-inch}$ board and swap registers with 96 [x⇄y] - $(T: 0, Z: 0, Y: 0, X: 3\ 1/2)$ to find the exact $3\ \frac{1}{2}\text{-inch}$ offcut!

Workbook Problem 5.20: Dividing Ceramic Frame Polish into Restoration Bottles

You have $\frac{2}{3}\text{ liter}$ of ceramic frame polish, and each full bicycle restoration uses $\frac{1}{4}\text{ liter}$. Why does dividing by $\frac{1}{4}$ make the number get bigger—and how do we model it with the exact same tower-height meaning as whole-number division? When we divide by a whole number (like $\frac{3}{4} \div 3 = \frac{1}{4}$), we ask: "If $\frac{3}{4}$ is $3$ equal parts of a target tower, how tall is $1$ full target tower?" Ask the exact same Split & Rebuild (1 Full Tower) question for $\frac{2}{3} \div \frac{1}{4}$: "If two $1/3$ blocks ($56\text{ mm}$) are just $\frac{1}{4}$ of a mystery tower, how tall is all $\frac{4}{4}$ ($1$ full mystery tower)?" Pooling blocks across your 4-kit classroom table pod, stack $4$ sets of two $1/3$ blocks ($4 \times 56\text{ mm} = 224\text{ mm} = \text{eight } 1/3\text{ blocks}$), which stands flush beside $2$ Whole blocks plus two $1/3$ blocks ($2\ \frac{2}{3}$)! (You can also verify by measurement in the Comparator Well: two $1/3$ blocks at $56\text{ mm}$ fit two $1/4$ blocks at $42\text{ mm}$ plus an extra $14\text{ mm}$, which is $\frac{2}{3}$ of a $21\text{ mm}$ quarter-block): [.] 2 [.] 3 [ENTER] [.] 1 [.] 4 [÷] [a ᵇ/c] $\rightarrow$ 2 2/3.

Pour the available polish . 2 . 3 into $Y$ with [ENTER] $(T: 0, Z: 0, Y: 2/3, X: 2/3)$, divide by the per-bike dose . 1 . 4 [÷] $(T: 0, Z: 0, Y: 0, X: 8/3)$, and unpack via [a ᵇ/c] $(T: 0, Z: 0, Y: 0, X: 2\ 2/3)$ to see $2$ full bike restorations plus $\frac{2}{3}$ of a third!

Workbook Problem 5.21: Snipping Braided Shifter Housing into Derailleur Loops

A $4\ \frac{1}{2}\text{-foot}$ coil of braided shifter cable housing must be snipped into $\frac{3}{4}\text{-foot}$ loops for rear derailleur bends. With your 4-kit table pod, you can see why dividing by $\frac{3}{4}$ always splits by the numerator ($\div 3$) and rebuilds by the denominator ($\times 4$): for instance, in $\frac{1}{2} \div \frac{3}{4}$, a $1/2$ block ($42\text{ mm}$) is $3$ fourths of a mystery tower, so splitting by $3$ gives one $1/6$ block ($14\text{ mm}$) per fourth, and stacking $4$ of those $1/6$ blocks ($56\text{ mm}$) builds the full mystery tower of $\frac{2}{3}$! Applying the same $(\div 3) \times 4$ rule to $4\ \frac{1}{2} \div \frac{3}{4}$: splitting $4\ \frac{1}{2}$ into $3$ equal parts gives $1\ \frac{1}{2}$, and multiplying by $4$ gives $6$. Confirm how many loops come out of the coil on your stack: [4] [.] 1 [.] 2 [ENTER] [.] 3 [.] 4 [÷] $\rightarrow$ 36/6.

Place the coil length 4 . 1 . 2 into $Y$ with [ENTER] $(T: 0, Z: 0, Y: 4\ 1/2, X: 4\ 1/2)$ and divide by the loop length . 3 . 4 [÷] $(T: 0, Z: 0, Y: 0, X: 6)$ to get $6$ derailleur loops with zero waste.

Workbook Problem 5.22: The Domino Chain Inside a Three-Stage Planetary Hub

Inside a three-stage bicycle hub, stage 1 steps rotation down by $\frac{1}{2}$, stage 2 by $\frac{2}{3}$, and stage 3 by $\frac{3}{4}$. Look at the numerators and denominators lined up in a row: $\frac{1}{\mathbf{2}} \times \frac{\mathbf{2}}{\mathbf{3}} \times \frac{\mathbf{3}}{4}$—the inner $2$s and $3$s topple like dominoes, leaving just the outer $\frac{1}{4}$! Watch the SC-6 multiply the raw stages first and then collapse the whole chain with [g] [SIMP]: [.] 1 [.] 2 [ENTER] [.] 2 [.] 3 [×] [.] 3 [.] 4 [×] [g] [SIMP] $\rightarrow$ 1/4.

Start with stage 1 . 1 . 2 and [ENTER] $(T: 0, Z: 0, Y: 1/2, X: 1/2)$, chain stage 2 . 2 . 3 [×] $(T: 0, Z: 0, Y: 0, X: 2/6)$ and stage 3 . 3 . 4 [×] $(T: 0, Z: 0, Y: 0, X: 6/24)$, and tap [g] [SIMP] $(T: 0, Z: 0, Y: 0, X: 1/4)$ to reveal the telescoping $1/4$ hub ratio!

Workbook Problem 5.23: Clipping a Coupon on a Cozy Wool Work Coat

At the farm apparel stall, a heavy wool work coat costs $\$85.00$, and you have a $20\%$ off coupon. Since $20\%$ is $\frac{1}{5}$ of the price ($\$85 \div 5 = \$17$), the coupon knocks $\$17.00$ off the tag. Because [%] keeps $\$85.00$ waiting in $Y$, one press of [-] subtracts the discount automatically: 85 [ENTER] 20 [%] $\longrightarrow X = \$17.00$ discount. Press [-] $\longrightarrow X = \$68.00$ sale price.

Load the coat price 85 [ENTER] $(T: 0, Z: 0, Y: 85, X: 85)$, find the coupon savings 20 [%] $(T: 0, Z: 0, Y: 85, X: 17.00)$, and tap [-] $(T: 0, Z: 0, Y: 0, X: 68.00)$ to deduct $\$17.00$ from the waiting $\$85.00$ in a single stroke.

Workbook Problem 5.24: Ringing Up Sales Tax on the Heavy-Duty Garden Cart

Purchasing a $\$120.00$ heavy-duty market garden cart adds an $8.5\%$ local sales tax. Every $\$100$ of price adds $\$8.50$ of tax, and the extra $\$20$ (one-fifth of $\$100$) adds $\frac{1}{5} \times \$8.50 = \$1.70$, for $\$10.20$ in tax. Let Register $Y$ hold the $\$120.00$ base price while you compute and add the tax: 120 [ENTER] 8.5 [%] $\longrightarrow X = \$10.20$ tax. Press [+] $\longrightarrow X = \$130.20$ total.

Park the cart price 120 [ENTER] $(T: 0, Z: 0, Y: 120, X: 120)$, compute the tax 8.5 [%] $(T: 0, Z: 0, Y: 120, X: 10.20)$, and press [+] $(T: 0, Z: 0, Y: 0, X: 130.20)$ to fuse price and tax into $\$130.20$.

Workbook Problem 5.25: Grading Golden Honey Jars at the Inspection Booth

At the artisan honey booth, quality control holds jars up to the sunlight: $42$ out of $50$ jars pass the golden clarity test. Because per-cent literally means "out of $100$," and $50$ is half of $100$, doubling both numbers ($42 \times 2 = 84$ out of $100$) predicts $84\%$ right away! Confirm the exact rate on your SC-6: 42 [ENTER] 50 [÷] 100 [×].

Divide the passing jars 42 [ENTER] 50 [÷] $(T: 0, Z: 0, Y: 0, X: 0.84)$ to see the decimal rate $0.84$, then scale to parts-per-hundred with 100 [×] $(T: 0, Z: 0, Y: 0, X: 84)$ to confirm $84\%$.

Workbook Problem 5.26: End-of-Season Clearance on Trail Boots

A pair of leather trail boots originally priced at $\$110.00$ is marked $30\%$ off on the clearance table. With eleven $\$10$ bills, taking $30\%$ off removes $\$3$ from every $\$10$ bill ($11 \times \$3 = \$33$ saved), leaving $70\%$ of the price ($\$77.00$): 110 [ENTER] 30 [%] [-].

Compute the markdown with 110 [ENTER] 30 [%] $(T: 0, Z: 0, Y: 110, X: 33.00)$, then tap [-] $(T: 0, Z: 0, Y: 0, X: 77.00)$ to subtract the $\$33.00$ savings from the $\$110.00$ waiting in $Y$.

Workbook Problem 5.27: Counting Seats for the Harvest Co-Op Banquet

Of the $450$ members in the regional growers' co-op, $60\%$ RSVP for the annual harvest banquet. Splitting $450$ into ten equal tables gives $45$ members per $10\%$, so six tables hold $6 \times 45 = 270$ guests: 450 [ENTER] 60 [%].

Place the total membership 450 [ENTER] $(T: 0, Z: 0, Y: 450, X: 450)$ on the stack and tap 60 [%] $(T: 0, Z: 0, Y: 450, X: 270)$ to count $270$ banquet seats.

Workbook Problem 5.28: Stacking Two Discounts on a Solar Battery Station

At the market hardware depot, a $\$200.00$ solar backup battery is on sale for $20\%$ off, and co-op members get an extra $10\%$ off the reduced sale price. Why doesn't $20\%$ off followed by $10\%$ off equal $30\%$ off ($\$140$)? Because the second $10\%$ discount applies to the smaller $\$160$ sale price, saving $\$16$ instead of $\$20$! Watch both discounts chain cleanly on the stack: 200 [ENTER] 20 [%] [-] 10 [%] [-].

Take the first markdown 200 [ENTER] 20 [%] [-] $(T: 0, Z: 0, Y: 0, X: 160)$ to reach the $\$160$ sale price, then chain the member discount right onto that new base with 10 [%] [-] $(T: 0, Z: 0, Y: 0, X: 144.00)$ for a final checkout price of $\$144.00$.

Workbook Problem 5.29: Slowing Down the Drawbridge Winch Drum

The drawbridge motor spins at $1,800\text{ RPM}$. Because the $1:3$ gear ratio means the $12$-tooth pinion must spin $3$ full turns to rotate the $36$-tooth sector gear just once, the sector gear turns at one-third the speed (with $3$ times the lifting torque!): 1800 [ENTER] 3 [÷].

Divide the motor speed 1800 [ENTER] 3 [÷] $(T: 0, Z: 0, Y: 0, X: 600)$ to see the winch drum turn at a steady $600\text{ RPM}$.

Workbook Problem 5.30: Mixing the Living-Wall Hydroponic Reservoir

Mixing liquid plant food for an automated living-wall model blends $2$ parts nutrient concentrate with $15$ parts water to fill a $34\text{-liter}$ reservoir. Here is the secret to every part-to-part mixture puzzle: add the parts together first ($2 + 15 = 17\text{ total parts}$) to find how much liquid goes into a single part ($34 \div 17 = 2\text{ liters per part}$)! Keystrokes: 1. Total parts: 2 [ENTER] 15 [+] $\longrightarrow 17$ parts. 2. Volume per part: 34 [ENTER] 17 [÷] $\longrightarrow 2.0\text{ L/part}$. 3. Concentrate: 2 [ENTER] 2 [×] $\longrightarrow 4.0\text{ L}$. 4. Water: 15 [ENTER] 2 [×] $\longrightarrow 30.0\text{ L}$.

Sum the recipe parts 2 [ENTER] 15 [+] $(T: 0, Z: 0, Y: 0, X: 17)$, then divide into the $34\text{-liter}$ tank via 34 [x⇄y] ÷ $(T: 0, Z: 0, Y: 0, X: 2)$ to find $2\text{ L/part}$—giving $4.0\text{ L}$ of concentrate and $30.0\text{ L}$ of water ($4 + 30 = 34\text{ L}$)!

Workbook Problem 5.31: Scooping Cranberry Trail Mix for the Drafting Crew

During late-afternoon drafting sessions in the model shop, you mix $16\text{ cups}$ of trail mix made of $\frac{3}{8}$ dried cranberries, $\frac{1}{4}$ roasted almonds, and $\frac{3}{8}$ dark chocolate drops. Notice how $\frac{3}{8} + \frac{2}{8} + \frac{3}{8} = \frac{8}{8}$ accounts for the whole bowl! Since $\frac{1}{8}$ of $16\text{ cups}$ is $2\text{ cups}$, three eighths must be $3 \times 2 = 6\text{ cups}$: 16 [ENTER] [.] 3 [.] 8 [×] $\rightarrow$ 48/8.

Put the batch size 16 [ENTER] $(T: 0, Z: 0, Y: 16, X: 16)$ on the stack and multiply by the cranberry share . 3 . 8 [×] $(T: 0, Z: 0, Y: 0, X: 6)$ to scoop out $6\text{ cups}$ of cranberries.

Workbook Problem 5.32: Balancing Base Color and White Tint for Gallery Stucco

Mixing custom facade stucco pigment for a gallery model combines $3\ \frac{1}{2}\text{ pints}$ of terracotta base color with $1\ \frac{3}{4}\text{ pints}$ of white tint. Look closely at those two numbers in fourths: $3\ \frac{1}{2} = \frac{14}{4}$ is exactly double $1\ \frac{3}{4} = \frac{7}{4}$, so your stucco follows a clean $2:1$ color-to-tint ratio! Sum the two volumes on the stack to check the total jar size needed: 3 [.] 1 [.] 2 [ENTER] 1 [.] 3 [.] 4 [+] [a ᵇ/c] $\rightarrow$ 5 1/4.

Pour in the base pigment 3 . 1 . 2 with [ENTER] $(T: 0, Z: 0, Y: 3\ 1/2, X: 3\ 1/2)$, blend the white tint 1 . 3 . 4 [+] $(T: 0, Z: 0, Y: 0, X: 21/4)$, and unpack with [a ᵇ/c] $(T: 0, Z: 0, Y: 0, X: 5\ 1/4)$ to fill a $5\ \frac{1}{4}\text{-pint}$ container.

Workbook Problem 5.33: Scaling the Light-Rail Blueprint Up to Real Miles

On a regional topographical blueprint, $1\text{ inch}$ represents $3\ \frac{1}{2}\text{ miles}$, and the proposed light-rail corridor measures $4\ \frac{1}{4}\text{ inches}$ long. You can estimate first: $4\text{ inches}$ is $4 \times 3\ \frac{1}{2} = 14\text{ miles}$, plus a quarter of $3\ \frac{1}{2}\text{ miles}$ ($\frac{7}{8}\text{ mile}$). Multiply the two mixed numbers on the stack to confirm: 4 [.] 1 [.] 4 [ENTER] 3 [.] 1 [.] 2 [×] [a ᵇ/c] $\rightarrow$ 14 7/8.

Place the blueprint length 4 . 1 . 4 into $Y$ with [ENTER] $(T: 0, Z: 0, Y: 4\ 1/4, X: 4\ 1/4)$, scale by the map factor 3 . 1 . 2 [×] $(T: 0, Z: 0, Y: 0, X: 119/8)$, and tap [a ᵇ/c] $(T: 0, Z: 0, Y: 0, X: 14\ 7/8)$ to verify the $14\ \frac{7}{8}\text{-mile}$ transit line.

Workbook Problem 5.34: Clocking Four Loops on the Miniature Tramway

Timing four automated scale tram loops on a miniature transit layout records split times of $1\ \frac{1}{10}$, $1\ \frac{1}{6}$, $1\ \frac{1}{12}$, and $1\ \frac{1}{15}$ minutes. Why do those four strange denominators ($10, 6, 12,$ and $15$) play together so nicely on a stopwatch? Because all four divide evenly into $60$ seconds ($6\text{ s} + 10\text{ s} + 5\text{ s} + 4\text{ s} = 25\text{ s}$, or $\frac{25}{60} = \frac{5}{12}\text{ minute}$)! Chain all four splits on the stack: 1 [.] 1 [.] 10 [ENTER] 1 [.] 1 [.] 6 [+] 1 [.] 1 [.] 12 [+] 1 [.] 1 [.] 15 [+] [a ᵇ/c] $\rightarrow$ 4 5/12.

Start with loop 1 1 . 1 . 10 $(T: 0, Z: 0, Y: 0, X: 1\ 1/10)$, accumulate loop 2 1 . 1 . 6 [+] $(T: 0, Z: 0, Y: 0, X: 34/15)$, loop 3 1 . 1 . 12 [+] $(T: 0, Z: 0, Y: 0, X: 67/20)$, and loop 4 1 . 1 . 15 [+] $(T: 0, Z: 0, Y: 0, X: 53/12)$, then tap [a ᵇ/c] $(T: 0, Z: 0, Y: 0, X: 4\ 5/12)$ to read $4\ \frac{5}{12}\text{ minutes}$ ($4\text{ min } 25\text{ s}$)!

Workbook Problem 5.35: Dueling Copper-Foil Suppliers on Two Stack Rows

In the Science Olympiad circuit-building event, Supplier A sells $18\text{ oz}$ of adhesive copper foil for $\$3.24$, while Supplier B sells $24\text{ oz}$ for $\$4.08$. Because the SC-6 displays Register $Y$ and Register $X$ simultaneously, you can compute Supplier A's unit price and then compute Supplier B's unit price right below it—letting both unit prices sit side-by-side on screen! Keystrokes: - Brand A: 3.24 [ENTER] 18 [÷] $\longrightarrow X = \$0.18/\text{oz}$. - Brand B: 4.08 [ENTER] 24 [÷] $\longrightarrow X = \$0.17/\text{oz}$.

Find Supplier A's unit cost with 3.24 [ENTER] 18 [÷] $(T: 0, Z: 0, Y: 0, X: 0.18)$, then compute Supplier B's right underneath via 4.08 [ENTER] 24 [÷] $(T: 0, Z: 0, Y: 0.18, X: 0.17)$—comparing $Y = \$0.18/\text{oz}$ against $X = \$0.17/\text{oz}$ at a glance!

Workbook Problem 5.36: Three-Second Tower Reduction in the Lightning Round

In the rapid-fire reduction sprint, you have 3 seconds to reduce $\frac{84}{144}$ to lowest terms. Notice those two numbers: $84\text{ mm}$ is the height of 1 Whole Fraction Studio tower ($7 \times 12$), and $144$ is a dozen dozens ($12 \times 12$)! Tap [g] [SIMP] to factor out $12$ in a single heartbeat: [.] 84 [.] 144 [ENTER] [g] [SIMP] $\rightarrow$ 7/12.

Key in . 84 . 144 $(T: 0, Z: 0, Y: 0, X: 84/144)$, preserve the original fraction in $Y$ with [ENTER] $(T: 0, Z: 0, Y: 84/144, X: 84/144)$, and tap [g] [SIMP] $(T: 0, Z: 0, Y: 84/144, X: 7/12)$ to see $\frac{84}{144}$ and its simplified form $\frac{7}{12}$ side-by-side!

Workbook Problem 5.37: Scaling Both Sides of a Gear Ratio in One Stroke

In the engineering proportions sprint, you need to scale a miniature $4:7$ gear ratio up by a factor of $15$ so the teeth are large enough to 3D-print. In Ratio Mode, multiplying by $15$ scales both sides of the colon simultaneously ($4 \times 15 = 60$ and $7 \times 15 = 105$): [.] 4 [.] 7 [g] [a:b] 15 [×] $\rightarrow$ 60:105.

Lock . 4 . 7 into Ratio Mode $(T: 0, Z: 0, Y: 0, X: 4:7)$ and scale by 15 [×] $(T: 0, Z: 0, Y: 0, X: 60:105)$ to get the $60:105$ tooth pair in one operation.

Workbook Problem 5.38: Proving the Secret Partner Rule Under the Tournament Clock

In the final number theory sprint, prove the Secret Partner Rule $\text{GCF}(36, 48) \times \text{LCM}(36, 48) = 36 \times 48$ live on the stack! Why does this always work? Because $\text{GCF}(36, 48) = 12$ collects the shared prime factors while $\text{LCM}(36, 48) = 144$ collects the full prime union—so multiplying them puts every single prime factor of $36$ and $48$ back together ($12 \times 144 = 1,728$): Keystrokes: 1. GCF: 36 [ENTER] 48 [GCF] $\longrightarrow 12$. 2. LCM: 36 [ENTER] 48 [f] [LCM] $\longrightarrow 144$. 3. Product: 12 [ENTER] 144 [×] $\longrightarrow 1,728$. 4. Direct: 36 [ENTER] 48 [×] $\longrightarrow 1,728$.

Extract the shared factor 36 [ENTER] 48 [GCF] $(T: 0, Z: 0, Y: 0, X: 12)$, compute the common multiple 36 [ENTER] 48 [f] [LCM] $(T: 0, Z: 0, Y: 12, X: 144)$ while $12$ waits in $Y$, and tap [×] $(T: 0, Z: 0, Y: 0, X: 1728)$ to prove $12 \times 144 = 1,728 = 36 \times 48$!

Chapter 06: Geometry, Surface Area & Planetary Geared Time — Extended Worked Problems

Accompanies Chapter 06 of From Blocks to the Stack (33 Worked Practice Problems)

Workbook Problem 6.1: Triangular Park Pavilion Pennant

A ceremonial triangular felt pennant hangs above the park entrance pavilion, with base $b = 25\text{ cm}$ along the crossbeam and vertical altitude $h = 60\text{ cm}$. Calculate the felt fabric required. Stretch a rubber band on your Geoboard into a rectangle and divide it diagonally: two identical right triangles emerge, showing that the area of one triangle is half of the enclosing rectangle ($A = \frac{1}{2} b h$). Evaluate the pennant area with [g] [TRI]: 25 [ENTER] 60 [g] [TRI] $\rightarrow$ 750 (T: 0, Z: 0, Y: 0, X: 750) to halve the $25 \times 60 = 1,500\text{ cm}^2$ bounding rectangle into $\mathbf{750.00\text{ cm}^2}$ of felt.

Workbook Problem 6.2: Floral Plaza Circular Sundial Clockface

A circular sundial in the central plaza has a diameter of $30\text{ cm}$ (radius $r = 15\text{ cm}$). The bronze faceplate must be cut from polished sheet metal. Inspect one of the circular end disks from your Cylindrical Wrapper Net: slicing a circle into wedge sectors and alternating them forms a parallelogram of height $r$ and base $\pi r$, giving area $\pi r^2$. Compute the disk area using [f] [CIRC]: 15 [f] [CIRC] $\rightarrow$ 706.85834706 (T: 0, Z: 0, Y: 0, X: 706.8583) to scale the $15^2 = 225\text{ cm}^2$ radius square by $\pi$, yielding $\mathbf{706.86\text{ cm}^2}$ of bronze.

Workbook Problem 6.3: Parallelogram Flower Bed

An ornamental botanical flower bed is laid out as a parallelogram with base $b = 8\text{ meters}$ and perpendicular altitude $h = 4.5\text{ meters}$. Calculate the topsoil area. On your Geoboard, form a parallelogram and notice that slicing off the right triangle on one end and sliding it to the other end forms a rectangle of identical base and height ($A = b \times h$). Multiply base by height on the SC-6: [8] [ENTER] 4.5 [×] $\rightarrow$ 36 (T: 0, Z: 0, Y: 0, X: 36), proving the sheared bed holds the exact same $\mathbf{36.00\text{ m}^2}$ area as an $8 \times 4.5\text{ m}$ rectangle.

Workbook Problem 6.4: Rhombus Decorative Pathway Pavers

Plaza pathways are tiled with diamond-shaped granite pavers having diagonal spans $d_1 = 40\text{ cm}$ and $d_2 = 70\text{ cm}$. Find the paver face area ($A = \frac{1}{2} d_1 d_2$). Stretch a diamond (rhombus) on your Geoboard: its perpendicular diagonals split the rhombus into four right triangles that fit inside a bounding rectangle of area $d_1 \times d_2$, occupying half its area. Compute the rhombus area on the SC-6: 40 [ENTER] 70 [×] [2] [÷] $\rightarrow$ 1400.

Multiply the two diagonals 40, [ENTER], 70, and [×] to frame the $2,800\text{ cm}^2$ bounding box around the diamond (T: 0, Z: 0, Y: 0, X: 2800), then fold the four outer corner triangles inward with 2 and [÷] (T: 0, Z: 0, Y: 0, X: 1400) to reveal a paver face area of $\mathbf{1400.00\text{ cm}^2}$.

Workbook Problem 6.5: Irrigation Pipe Annular Gasket Ring

The park irrigation line uses neoprene annular washers with outer radius $R = 25\text{ mm}$ and inner pipe cutout radius $r = 10\text{ mm}$. Calculate the rubber seal contact area ($A = \pi (R^2 - r^2)$). Removing a $10\text{ mm}$ inner disk from a $25\text{ mm}$ outer circle leaves a ring whose area equals the difference between the outer and inner circle areas. Calculate the ring area on the SC-6: 25 [x²] 10 [x²] [-] [π] [×] $\rightarrow$ 1,649.336143.

Subtract the inner square from the outer square (25, [x²] for $625$, then 10, [x²] for $100$, and [-]) to get $R^2 - r^2 = 525\text{ mm}^2$ before touching $\pi$ (T: 0, Z: 0, Y: 0, X: 525), then multiply by [π] and [×] once at the end (T: 0, Z: 0, Y: 0, X: 1649.3361) for a gasket contact area of $\mathbf{1649.34\text{ mm}^2}$.

Workbook Problem 6.6: Thermal Foil for a Modular Cube Telemetry Pod

A miniature cubic GPS beacon deployed on the Ross Ice Shelf measures $s = 2\text{ cm}$ along each edge. Calculate its total six-sided surface area. Examine a unit cube from your Place Value kit (or a 3-tile stack of $19.0 \times 19.0\text{ mm}$ Counting Tiles): a cube has 6 identical square faces, so calculating one square face ($s^2$) and multiplying by 6 yields the complete outer skin. Compute the 6-face surface area on the SC-6: [2] [x²] [6] [×] $\rightarrow$ 24 (T: 0, Z: 0, Y: 0, X: 24), squaring the $2\text{ cm}$ edge ($4\text{ cm}^2$) and multiplying by 6 facets for a beacon surface area of $\mathbf{24.00\text{ cm}^2}$.

Workbook Problem 6.7: Lateral Surface Area of a Conical Field Radar Beacon Cap

An ice-field radar transponder is protected by a bright orange conical fiberglass cowl with base radius $r = 8\text{ cm}$ and slant height $s = 18\text{ cm}$. The base is open to mount onto the antenna mast. Find lateral fiberglass surface area ($A = \pi r s$). Unrolling the curved wall of a cone along its slant seam forms a circular sector of radius $s$ and arc length $2\pi r$, whose area simplifies to $\pi r s$. Multiply $\pi \times 8 \times 18$ on the stack: [π] [8] [×] 18 [×] $\rightarrow$ 452.38934212.

Form the half-circumference arc [π], 8, and [×] ($25.1327\text{ cm}$) (T: 0, Z: 0, Y: 0, X: 25.1327), then sweep it along the slant height 18 and [×] (T: 0, Z: 0, Y: 0, X: 452.3893) to wrap the radar cowl in $\mathbf{452.39\text{ cm}^2}$ of fiberglass.

Workbook Problem 6.8: Open-Top Cold-Storage Battery Cargo Pod

A rugged open-top insulated cargo bin for lithium power cells measures length $L = 30\text{ cm}$, width $W = 20\text{ cm}$, and depth $H = 15\text{ cm}$. Calculate interior protective lining area across its 5 open-top faces (bottom floor, two side walls, two end walls). Unfold the Rectangular Prism Net and fold back the top lid panel: the remaining 5 faces form a cross with 1 central floor ($30 \times 20$) and 4 perimeter walls. Sum the 5 open-top faces with [Σ+]: [g] [CLΣ] 30 [ENTER] 20 [×] [Σ+] 30 [ENTER] 15 [×] [2] [×] [Σ+] 20 [ENTER] 15 [×] [2] [×] [Σ+] [f] [Σx] $\rightarrow$ 2100.

Clear the accumulator with [g] [CLΣ] (T: 0, Z: 0, Y: 0, X: 0), lay the single bottom floor (30, [ENTER], 20, [×]) of $600\text{ cm}^2$ (T: 0, Z: 0, Y: 0, X: 600) into [Σ+] (T: 0, Z: 0, Y: 0, X: 1), add the twin long walls (30, [ENTER], 15, [×] for $450\text{ cm}^2$, doubled by 2 and [×] to $900\text{ cm}^2$) with [Σ+] (T: 0, Z: 0, Y: 0, X: 2), add the twin end walls (20, [ENTER], 15, [×] for $300\text{ cm}^2$, doubled by 2 and [×] to $600\text{ cm}^2$) with [Σ+] (T: 0, Z: 0, Y: 0, X: 3), and tap [f] [Σx] (T: 0, Z: 0, Y: 0, X: 2100) to sum all 5 open-top faces to $\mathbf{2100.00\text{ cm}^2}$.

Workbook Problem 6.9: Hollow Ice Core Borehole Thermal Sleeve

A vertical borehole casing sleeve drilled into the polar ice sheet has inner radius $r = 0.5\text{ m}$ and depth $L = 4\text{ m}$. Find the inner cylindrical surface area lining the ice borehole. A hollow open-ended cylinder has no top or bottom caps; its surface is simply the unrolled rectangle of the Cylindrical Wrapper Net, equal to circumference times depth ($A = 2 \pi r L$). Compute the lateral sleeve area on the SC-6: 0.5 [ENTER] [2] [×] [π] [×] [4] [×] $\rightarrow$ 12.566370614.

Double the radius (0.5, [ENTER], 2, [×]) to get the $1.0\text{ m}$ borehole diameter (T: 0, Z: 0, Y: 0, X: 1), wrap around the rim with [π] and [×] for a $3.1416\text{ m}$ circumference (T: 0, Z: 0, Y: 0, X: 3.1416), and extrude down the 4 and [×] depth (T: 0, Z: 0, Y: 0, X: 12.5664) to line $\mathbf{12.57\text{ m}^2}$ of glacial ice.

Workbook Problem 6.10: Deep-Freeze Survival Rations Container Volume

Emergency freeze-dried meal bricks are packed into a heavy-duty container measuring length $18\text{ cm}$, width $6\text{ cm}$, and height $28\text{ cm}$. Find the total internal volume in cubic centimeters. With your kit tiles, lay a miniature $3 \times 1$ base (3 tiles) and stack 2 layers high ($3 \times 1 \times 2 = 6\text{ tiles}$) to see how layers build 3D volume ($V = l \times w \times h$). Now calculate the full $18 \times 6 \times 28\text{ cm}$ rations container volume on the SC-6: 18 [ENTER] [6] [×] 28 [×] $\rightarrow$ 3024.

Tile the container floor (18, [ENTER], 6, [×]) to form a $108\text{ cm}^2$ base layer (T: 0, Z: 0, Y: 0, X: 108), then stack 28 layers high with [×] (T: 0, Z: 0, Y: 0, X: 3024) to pack $\mathbf{3024.00\text{ cm}^3}$ of survival rations.

Workbook Problem 6.11: Modular Polar Research Station Habitat Block Volume

An insulated habitat module is assembled from 24 prefabricated cubic units, each measuring $2\text{ cm} \times 2\text{ cm} \times 2\text{ cm}$ on a 1:100 engineering architectural model. Calculate total module volume. Using 8 unit cubes from your Place Value kit, assemble a $2 \times 2 \times 2$ cube to see that $2^3 = 8$ cubic units, then multiply by 24 modules on the stack: [2] [ENTER] [3] [g] [yˣ] 24 [×] $\rightarrow$ 192.

Cube the module edge (2, [ENTER], 3, [g] [yˣ]) to see why doubling a $1\text{ cm}$ cube's edge multiplies its volume by $8$ (T: 0, Z: 0, Y: 0, X: 8), then multiply by 24 and [×] (T: 0, Z: 0, Y: 0, X: 192) for $\mathbf{192.00\text{ cm}^3}$ across the entire station.

Workbook Problem 6.12: Conical Sand Filtration Drainage Pile Volume

A sand filtration pile at the municipal water treatment facility forms a cone with base radius $r = 3\text{ meters}$ and vertical peak height $h = 2\text{ meters}$. Find the total volume of filter sand. Recall the $3:1$ pouring relationship between a cone and a cylinder of the same base and height: three cones fill one cylinder, so a cone's volume is $\frac{1}{3}\pi r^2 h$. Compute the conical volume on the stack: [3] [f] [CIRC] [2] [×] [3] [÷] $\rightarrow$ 18.849555922.

Build the enclosing cylinder first (3, [f] [CIRC] for $28.2743\text{ m}^2$, times height 2 and [×] for $56.5487\text{ m}^3$) (T: 0, Z: 0, Y: 0, X: 56.5487), then take one-third via 3 and [÷] (T: 0, Z: 0, Y: 0, X: 18.8496) to prove the sand cone holds $\mathbf{18.85\text{ m}^3}$.

Workbook Problem 6.13: Square Pyramid Reservoir Enclosure Roof

A square pyramid roof caps an outdoor water storage reservoir. The base measures $s = 12\text{ m}$ on each side, and the vertical peak height is $h = 5\text{ m}$. Find the enclosed roof volume. Fold the Square Pyramid Net from your kit into a 3D pyramid: its four triangular walls slant inward to a single apex, enclosing one-third the volume of a $12 \times 12 \times 5\text{ m}$ rectangular prism. Calculate the pyramid volume on the SC-6: 12 [x²] [5] [×] [3] [÷] $\rightarrow$ 240.

Square the roof footprint with 12 and [x²] ($144\text{ m}^2$) (T: 0, Z: 0, Y: 0, X: 144), raise the bounding box by 5 and [×] ($720\text{ m}^3$), and apply the $3:1$ pyramid taper with 3 and [÷] (T: 0, Z: 0, Y: 0, X: 240) to enclose $\mathbf{240.00\text{ m}^3}$.

Workbook Problem 6.14: Giant Science Museum Soap-Bubble Sphere Volume

At the Science Center's bubble lab, you blow a giant spherical soap bubble with radius $r = 12\text{ cm}$ ($24\text{ cm}$ across). Find the total volume of air trapped inside the sphere. By Archimedes' theorem, a sphere occupies two-thirds the volume of its enclosing cylinder, giving $V = \frac{4}{3}\pi r^3$. Evaluate the spherical volume on the stack: [4] [ENTER] [3] [÷] [π] [×] 12 [ENTER] [3] [g] [yˣ] [×] $\rightarrow$ 7,238.229474.

Stage Archimedes' spherical constant with 4 [ENTER] 3 [÷] [π] [×] ($\frac{4}{3}\pi \approx 4.1888$), cube the bubble radius via 12 [ENTER] 3 [g] [yˣ] ($1,728\text{ cm}^3$) (T: 0, Z: 0, Y: 4.1888, X: 1728), and fuse them with [×] (T: 0, Z: 0, Y: 0, X: 7238.2295) to trap $\mathbf{7238.23\text{ cm}^3}$ of shimmering air.

Workbook Problem 6.15: Hemispherical Planetarium Star-Dome Volume

The Science Center's kids' planetarium features a hemispherical star-projection dome with radius $r = 8\text{ meters}$. Find the total enclosed air volume beneath the dome. Because a hemisphere is half of a full sphere, compute the full sphere volume $\frac{4}{3}\pi r^3$ and divide by 2 on the stack: [4] [ENTER] [3] [÷] [π] [×] [8] [ENTER] [3] [g] [yˣ] [×] [2] [÷] $\rightarrow$ 1,072.330292.

Inflate the full $8\text{ m}$ sphere first with 4 [ENTER] 3 [÷] [π] [×] 8 [ENTER] 3 [g] [yˣ] [×] ($2,144.6606\text{ m}^3$) (T: 0, Z: 0, Y: 0, X: 2144.6606), then slice it along the horizon with 2 [÷] (T: 0, Z: 0, Y: 0, X: 1072.3303) to reveal the star-dome volume of $\mathbf{1072.33\text{ m}^3}$.

Workbook Problem 6.16: Eureka! Mystery Geode Rock Water Displacement

In the geology lab, you drop an irregular crystal geode rock into a graduated cylinder of radius $r = 2.5\text{ cm}$. The water level rises by $\Delta h = 3.2\text{ cm}$. Just like Archimedes in his bathtub, find the exact volume of the rock! Because the displaced water forms a cylindrical layer of radius $2.5\text{ cm}$ and height $3.2\text{ cm}$, its volume equals the rock's volume. Compute the displaced cylinder volume on the SC-6: 2.5 [f] [CIRC] 3.2 [×] $\rightarrow$ 62.831853072.

Find the beaker's circular cross-section with 2.5 and [f] [CIRC] ($19.6350\text{ cm}^2$) (T: 0, Z: 0, Y: 0, X: 19.635), then multiply by the 3.2 and [×] water rise (T: 0, Z: 0, Y: 0, X: 62.8319) to measure the jagged geode at $\mathbf{62.83\text{ cm}^3}$ ($62.83\text{ mL}$).

Workbook Problem 6.17: Science Museum Fountain Pump Shift Duration via `[f] [▸HR]`

The Science Center's musical water fountain starts at 8:15 AM ($8.15\text{ HMS}$) and pauses for afternoon inspection at 3:05 PM ($15.05\text{ HMS}$). Find total active fountain duration. Set the hands of your Planetary Geared Clock to 8:15 and rotate the minute ring forward to 3:05 (15:05): 6 full hours and 50 minutes elapse, demonstrating base-60 subtraction. Convert both times with [f] [▸HR] and subtract on the stack: 15.05 [f] [▸HR] 8.15 [f] [▸HR] [-] [g] [▸HMS] $\rightarrow$ 6.5.

Translate both base-60 timestamps into decimal hours and subtract (15.05 [f] [▸HR] 8.15 [f] [▸HR] [-]) to see why $15\frac{5}{60} - 8\frac{15}{60} = 6.8333\text{ hr}$ (T: 0, Z: 0, Y: 0, X: 6.8333), then snap back to clock minutes with [g] [▸HMS] (T: 0, Z: 0, Y: 0, X: 6.50) for an active shift of $\mathbf{6\text{ hr } 50\text{ min}}$ ($6.50\text{ HMS}$).

Workbook Problem 6.18: Aquarium Filter Cycle Completion Time

The giant kelp-forest aquarium filter cycle starts at 7:10 AM ($7.10\text{ HMS}$) and runs continuously for 3 hours and 48 minutes ($3.48\text{ HMS}$). Determine exact completion time. Turn your Planetary Geared Clock from 7:10 forward by 3 hours to 10:10, then advance 48 minutes to 10:58. Add the start time and runtime in decimal hours on the SC-6: 7.10 [f] [▸HR] 3.48 [f] [▸HR] [+] [g] [▸HMS] $\rightarrow$ 10.58.

Convert both clock readings to decimal hours and add (7.10 [f] [▸HR] 3.48 [f] [▸HR] [+]) to reach $10.9667\text{ hr}$ (T: 0, Z: 0, Y: 0, X: 10.9667), then tap [g] [▸HMS] (T: 0, Z: 0, Y: 0, X: 10.58) to see the cycle finish at $\mathbf{10:58\text{ AM}}$ ($10.58\text{ HMS}$).

Workbook Problem 6.19: Planetary Clock Hand Angle at 8:45

A cockpit heading dial displays flight waypoint time 8:45 ($8.45\text{ HMS}$). Determine continuous hour hand angle using $(8 + 45/60) \times 30^\circ$. Advance the Planetary Geared Clock to 8:45: the minute hand points at 9 ($270^\circ$), while the planetary gearing smoothly rotates the hour hand three-quarters past 8 to $262.5^\circ$. Calculate the exact angle on the stack: [8] [ENTER] 45 [ENTER] 60 [÷] [+] 30 [×] $\rightarrow$ 262.5.

Hold hour 8 in $Y$ with [ENTER] $(T: 0, Z: 0, Y: 8, X: 8)$, turn 45 minutes into three-quarters of an hour (45, [ENTER], 60, [÷]) $(T: 0, Z: 0, Y: 8, X: 0.75)$ and join via [+] $(T: 0, Z: 0, Y: 0, X: 8.75)$, then multiply by 30 and [×] $(T: 0, Z: 0, Y: 0, X: 262.5)$—proving the hour hand sits at $262.5^\circ$, just $7.5^\circ$ shy of the minute hand at $270^\circ$!

Workbook Problem 6.20: Angular Separation Between Clock Hands at 2:30

An azimuth radar scope aligns hands at 2:30. The minute hand points at $180^\circ$, and the continuous hour hand rotates to $(2 + 30/60) \times 30^\circ = 75^\circ$. Find the angular separation between hands. Set your Planetary Geared Clock to 2:30 and inspect the wedge between the minute hand ($180^\circ$) and hour hand ($75^\circ$). Subtract the two angles on the SC-6: 180 [ENTER] 2.5 [ENTER] 30 [×] [-] $\rightarrow$ 105.

Place the minute hand's straight-down angle 180 in $Y$ with [ENTER] $(T: 0, Z: 0, Y: 180, X: 180)$, find the hour hand's halfway position between 2 and 3 (2.5, [ENTER], 30, [×]) $(T: 0, Z: 0, Y: 180, X: 75)$, and take the difference with [-] $(T: 0, Z: 0, Y: 0, X: 105)$ to measure the $105^\circ$ wedge between the hands.

Workbook Problem 6.21: Epicyclic Planetary Gear Train Reduction Ratio

A mechanical aircraft instrument drive utilizes a planetary gear train with a sun gear of $N_s = 14$ teeth and an outer stationary ring gear of $N_r = 154$ teeth. Find the planetary reduction ratio $R = 1 + N_r / N_s$. Just like the $12:1$ gear train inside your Planetary Geared Clock, 12 full turns of the input drive rotate the output carrier through 1 complete turn. Verify the gear ratio on the SC-6: 154 [ENTER] 14 [÷] [1] [+] $\rightarrow$ 12.

Divide the outer ring teeth by the center sun teeth (154, [ENTER], 14, [÷]) to get the $11:1$ ring-to-sun ratio (T: 0, Z: 0, Y: 0, X: 11), then add the 1 and [+] planetary carrier revolution (T: 0, Z: 0, Y: 0, X: 12) to lock in an exact $\mathbf{12.0000:1}$ clockwork reduction!

Workbook Problem 6.22: Backyard Birdhouse Tower & Pyramid Roof Volume

You are building a deluxe wooden birdhouse shaped like a rectangular box base ($15 \times 15 \times 20\text{ cm}$) topped by a slanted square pyramid roof ($15 \times 15 \times 9\text{ cm}$). Calculate the total enclosed internal volume! Fold the Rectangular Prism Net and Square Pyramid Net from your kit and place the pyramid roof on top of the box base to see how the total volume combines the rectangular prism ($l \times w \times h$) with one-third of the roof's bounding box ($\frac{1}{3} s^2 h$). Sum both volumes on the stack: 15 [x²] 20 [×] 15 [x²] [9] [×] [3] [÷] [+] $\rightarrow$ 5175.

Build the lower nesting box first (15, [x²] for the $225\text{ cm}^2$ floor, times 20 and [×]) to hold $4,500\text{ cm}^3$ (T: 0, Z: 0, Y: 0, X: 4500), then build the pyramid attic right above it on the stack (15, [x²], 9, [×] for $2,025\text{ cm}^3$, tapered by 3 and [÷] to $675\text{ cm}^3$) (T: 0, Z: 0, Y: 4500, X: 675), and join box and roof with [+] (T: 0, Z: 0, Y: 0, X: 5175) for $\mathbf{5175.00\text{ cm}^3}$.

Workbook Problem 6.23: Saturday Pet-Sitting & Dog-Walking Shift Earnings

You help run a neighborhood Saturday dog-walking and pet-care service from 9:15 AM to 2:45 PM ($14.45\text{ HMS}$) at a team rate of $\$18.50/\text{hour}$. Find total earnings using [f] [▸HR]. Dial your Planetary Geared Clock from 9:15 to 2:45 (14:45): converting 9:15 AM to $9.25\text{ hr}$ and 2:45 PM to $14.75\text{ hr}$ shows an exact shift length of $5.5\text{ hours}$. Multiply the decimal hours by the hourly rate on the stack: 14.45 [f] [▸HR] 9.15 [f] [▸HR] [-] 18.50 [×] $\rightarrow$ 101.75.

Convert the afternoon finish 14.45 with [f] [▸HR] ($14.75\text{ hr}$) (T: 0, Z: 0, Y: 0, X: 14.75) and the morning start 9.15 with [f] [▸HR] ($9.25\text{ hr}$) (T: 0, Z: 0, Y: 14.75, X: 9.25), take the difference with [-] to get $5.5\text{ decimal hours}$ (T: 0, Z: 0, Y: 0, X: 5.5), and multiply directly by 18.50 and [×] (T: 0, Z: 0, Y: 0, X: 101.75) to earn $\mathbf{\$101.75}$.

Workbook Problem 6.24: Aircraft Emergency Water Ballast Pool Mass

A marine emergency flight test water ballast flume measures $25\text{ m} \times 10\text{ m} \times 2\text{ m}$. With fresh water density at $1,000\text{ kg/m}^3$, compute total water mass in kilograms. Build a scale model of the flume using 5 unit cubes from your Place Value kit ($5\text{ cm}^3$). Scaling up to the full pool, multiplying $25\text{ m} \times 10\text{ m} \times 2\text{ m}$ gives $500\text{ m}^3$, and multiplying by $1,000\text{ kg/m}^3$ yields total ballast mass: 25 [ENTER] 10 [×] [2] [×] 1000 [×] $\rightarrow$ 500000.

Spread the pool footprint (25, [ENTER], 10, [×]) to $250\text{ m}^2$ (T: 0, Z: 0, Y: 0, X: 250), fill it 2 meters deep with [×] for $500\text{ m}^3$ of water (T: 0, Z: 0, Y: 0, X: 500), and multiply by 1000 and [×] kilograms per cubic meter (T: 0, Z: 0, Y: 0, X: 500000) to reach $\mathbf{500,000\text{ kg}}$ ($500\text{ metric tons}$).

Workbook Problem 6.25: Flight Galley Conical Frustum Measuring Cup

A galley liquid measuring cup is engineered as a conical frustum with bottom radius $r = 3\text{ cm}$, top rim radius $R = 5\text{ cm}$, and depth $h = 7\text{ cm}$. Find fluid capacity in milliliters ($\text{cm}^3$) using $V = \frac{\pi h}{3}(R^2 + r^2 + Rr)$. Notice that a cylinder of radius $3\text{ cm}$ and height $7\text{ cm}$ would hold $\approx 198\text{ mL}$, while the flaring top rim expands the capacity to $359.19\text{ mL}$. Evaluate the frustum volume on the stack: [5] [x²] [3] [x²] [+] [5] [ENTER] [3] [×] [+] [π] [×] [7] [×] [3] [÷] $\rightarrow$ 359.

Sum the top and bottom radius squares (5, [x²] for $25$, plus 3, [x²] for $9$, [+]) to get $34$ (T: 0, Z: 0, Y: 0, X: 34), compute the geometric cross-product (5, [ENTER], 3, [×]) of $15$ (T: 0, Z: 0, Y: 34, X: 15) and blend with [+] into $R^2 + r^2 + Rr = 49$ (T: 0, Z: 0, Y: 0, X: 49), then scale by [π], [×] ($153.9380$), cup depth 7, [×] ($1077.5663$), and 3, [÷] (T: 0, Z: 0, Y: 0, X: 359.1888) for a capacity of $\mathbf{359.19\text{ mL}}$ ($359.19\text{ cm}^3$).

Workbook Problem 6.26: Aircraft Wing Core Hexagonal Prism Honeycomb Cell

A lightweight aircraft wing composite spar is built from regular hexagonal honeycomb cells with cell side $s = 3\text{ cm}$ and extrusion length $L = 10\text{ cm}$. Find the interior air volume of a single honeycomb chamber. Because a regular hexagon splits into six equilateral triangles of side $3\text{ cm}$, its base area is $\frac{3\sqrt{3}}{2}s^2$, and multiplying by length $L$ gives the prism volume. Calculate the honeycomb cell volume on the SC-6: [3] [x²] [3] [×] [3] [f] [√x] [×] [2] [÷] 10 [×] $\rightarrow$ 233.82685902.

Form the hexagonal factor 3, [x²] ($9$), 3, [×] for $3s^2 = 27$ (T: 0, Z: 0, Y: 0, X: 27), scale by the equilateral altitude ratio 3, [f] [√x] ($1.7321$), [×] ($46.7654$) (T: 0, Z: 0, Y: 0, X: 46.7654) and halve with 2, [÷] to get the $23.3827\text{ cm}^2$ hexagon footprint (T: 0, Z: 0, Y: 0, X: 23.3827), then extrude along the 10 and [×] spar depth (T: 0, Z: 0, Y: 0, X: 233.8269) for $\mathbf{233.83\text{ cm}^3}$.

Workbook Problem 6.27: Airfield Radar Tower Concrete Hollow Core Block

A foundation cinder block for a radar tower measures $40\text{ cm} \times 20\text{ cm} \times 20\text{ cm}$ externally, with twin interior rectangular voids measuring $14\text{ cm} \times 12\text{ cm} \times 20\text{ cm}$ each. Find net concrete volume. Model the hollow core concept on your desk using 16 unit cubes ($4 \times 2 \times 2$ block) and remove 4 center cubes. Then evaluate the full $40 \times 20 \times 20\text{ cm}$ envelope ($16,000\text{ cm}^3$) and subtract both $14 \times 12 \times 20\text{ cm}$ voids ($6,720\text{ cm}^3$) on the stack: 40 [ENTER] 20 [×] 20 [×] 14 [ENTER] 12 [×] 20 [×] [2] [×] [-] $\rightarrow$ 9280.

Build the outer block envelope (40, [ENTER], 20, [×] for $800\text{ cm}^2$, times height 20 and [×]) of $16,000\text{ cm}^3$ (T: 0, Z: 0, Y: 0, X: 16000), carve out the twin hollow cores right below it (14, [ENTER], 12, [×] for $168\text{ cm}^2$, times height 20, [×] and doubled by 2, [×]) to get $6,720\text{ cm}^3$ (T: 0, Z: 0, Y: 16000, X: 6720), and punch out the voids with [-] (T: 0, Z: 0, Y: 0, X: 9280) to leave $\mathbf{9280.00\text{ cm}^3}$ of solid concrete.

Workbook Problem 6.28: Navigational Dial Inscribed Equilateral Triangle Waste Area

A triangular compass pointer plate of side $s = 10\text{ cm}$ is laser cut from a circular acrylic disc of radius $R = \frac{s}{\sqrt{3}} \approx 5.7735\text{ cm}$. Find the waste acrylic scrap area ($A_{\text{waste}} = A_{\text{circle}} - A_{\text{triangle}}$). The three curved segments between the triangle's straight edges and the outer circular rim represent the leftover acrylic scrap. Subtract the triangle area from the circle area on the SC-6: 5.7735 [f] [CIRC] 10 [x²] [3] [f] [√x] [×] [4] [÷] [-] $\rightarrow$ 61.418387279.

Cut the circular acrylic blank with 5.7735 and [f] [CIRC] ($104.7197\text{ cm}^2$) (T: 0, Z: 0, Y: 0, X: 104.7197), compute the inscribed equilateral pointer (10, [x²] for $100$, times 3, [f] [√x], [×] for $173.2051$, divided by 4, [÷]) of $43.3013\text{ cm}^2$ (T: 0, Z: 0, Y: 104.7197, X: 43.3013), and lift out the pointer with [-] (T: 0, Z: 0, Y: 0, X: 61.4184) to find $\mathbf{61.42\text{ cm}^2}$ of leftover acrylic arcs.

Workbook Problem 6.29: Airport Weather Radar Geodesic Radome Strut Tubing

A 3V geodesic radar dome covering an airfield weather tracking array requires 30 Type-A struts ($1.20\text{ m}$), 40 Type-B struts ($1.35\text{ m}$), and 50 Type-C struts ($1.42\text{ m}$). Find total structural aluminum tubing length. Stretch rubber bands across pegs on your Geoboard to see how triangular struts lock together into a rigid truss, then multiply and sum the three strut batches ($30 \times 1.20\text{ m}$, $40 \times 1.35\text{ m}$, $50 \times 1.42\text{ m}$) directly on the stack: 30 [ENTER] 1.20 [×] 40 [ENTER] 1.35 [×] [+] 50 [ENTER] 1.42 [×] [+] $\rightarrow$ 161.

Tally the Type-A struts (30, [ENTER], 1.20, [×]) at $36\text{ m}$ (T: 0, Z: 0, Y: 0, X: 36), weld on the Type-B batch (40, [ENTER], 1.35, [×] for $54\text{ m}$, [+]) to reach $90\text{ m}$ (T: 0, Z: 0, Y: 0, X: 90), and add the Type-C batch (50, [ENTER], 1.42, [×] for $71\text{ m}$, [+]) (T: 0, Z: 0, Y: 0, X: 161) for $\mathbf{161.00\text{ meters}}$ of geodesic tubing.

Workbook Problem 6.30: Solar Noon to Sunset Daylight Duration Arc

On the equinox at the flight base, solar noon occurs at 12:00 PM and sunset occurs at 6:18 PM ($18.18\text{ HMS}$). Find total afternoon daylight duration as a continuous decimal hour. Turn your Planetary Geared Clock from 12:00 to 6:18 (18:18): converting 18 minutes past the hour gives $18/60 = 0.3\text{ hr}$, or $6.3\text{ decimal hours}$ of afternoon sun. Convert and subtract on the SC-6: 18.18 [f] [▸HR] 12 [-] $\rightarrow$ 6.3.

Convert sunset 18.18 with [f] [▸HR] to see why $18\text{ minutes}$ is $\frac{3}{10}$ of an hour ($18.3\text{ hr}$) (T: 0, Z: 0, Y: 0, X: 18.3), then subtract solar noon 12 with [-] (T: 0, Z: 0, Y: 0, X: 6.3) for $\mathbf{6.3000\text{ decimal hours}}$ ($6\text{ hr } 18\text{ min}$) of afternoon sun.

Workbook Problem 6.31: Submerged Bronze Sphere Hydrometer Water Rise

In the avionics fluids laboratory, a solid bronze calibration ball of radius $r = 3\text{ cm}$ is submerged into a cylindrical test beaker of radius $R = 5\text{ cm}$. Find the exact fluid height rise $\Delta h = \frac{V_{\text{sphere}}}{\pi R^2}$. Because the submerged sphere pushes aside its own volume of water ($V = \frac{4}{3}\pi r^3$), that displaced water spreads across the beaker's circular cross-section ($\pi R^2$) to raise the water level by $\Delta h = V / (\pi R^2)$. Evaluate the fluid rise on the stack: [4] [ENTER] [3] [÷] [π] [×] [3] [ENTER] [3] [g] [yˣ] [×] [5] [f] [CIRC] [÷] $\rightarrow$ 1.44.

Stage the sphere constant (4, [ENTER], 3, [÷], [π], [×]) of $4.1888$ (T: 0, Z: 0, Y: 0, X: 4.1888), multiply by the cubed ball radius (3, [ENTER], 3, [g] [yˣ] for $27\text{ cm}^3$, [×]) to find $113.0973\text{ cm}^3$ of displaced water (T: 0, Z: 0, Y: 0, X: 113.0973), then spread that water across the beaker's circular floor (5, [f] [CIRC] for $78.5398\text{ cm}^2$) (T: 0, Z: 0, Y: 113.0973, X: 78.5398) with [÷] (T: 0, Z: 0, Y: 0, X: 1.44) to watch the water level climb by $\mathbf{1.44\text{ cm}}$!

Workbook Problem 6.32: Rover Wheel Epicyclic Planetary Reduction

A robotic lunar sample rover wheel hub incorporates a planetary gear drive with sun gear $N_s = 10$, four planets $N_p = 25$, and stationary outer ring gear $N_r = 60$. Find the overall reduction ratio $R = 1 + N_r / N_s$. Just like the planetary gear train in your clock manipulative, every 7 turns of the central sun gear rotate the output carrier through 1 complete revolution. Compute the rover reduction ratio on the SC-6: 60 [ENTER] 10 [÷] [1] [+] $\rightarrow$ 7.

Divide the stationary ring teeth by the center sun teeth (60, [ENTER], 10, [÷]) to get $6$ (T: 0, Z: 0, Y: 0, X: 6), then add the 1 and [+] carrier orbit (T: 0, Z: 0, Y: 0, X: 7) to give the lunar rover a $\mathbf{7.0000:1}$ torque multiplication!

Workbook Problem 6.33: Flight Cargo Bullion Mass & Currency Valuation

High-value bullion cargo transport: A pure gold bar measures $15\text{ cm} \times 8\text{ cm} \times 4\text{ cm}$. With gold density at $19.3\text{ g/cm}^3$ and current market spot price at $\$65.00/\text{gram}$, calculate total ingot value using the SC-6 currency formatter [$]. Compare the rectangular prism volume ($480\text{ cm}^3$) to your folded $5 \times 4 \times 3\text{ cm}$ Rectangular Prism Net ($60\text{ cm}^3$): even though the gold bar is only 8 times the volume of your net, gold's high density ($19.3\text{ g/cm}^3$) packs $9.264\text{ kg}$ into that compact space. Chain volume, mass, and price on the stack: 15 [ENTER] [8] [×] [4] [×] 19.3 [×] 65 [×] [$] $\rightarrow$ $602160.00.

Lay the gold bar's base (15, [ENTER], 8, [×]) of $120\text{ cm}^2$ (T: 0, Z: 0, Y: 0, X: 120), stack it 4 cm thick with [×] for $480\text{ cm}^3$ (T: 0, Z: 0, Y: 0, X: 480), scale by gold's remarkable density 19.3 and [×] to hit $9,264\text{ grams}$ (T: 0, Z: 0, Y: 0, X: 9264), multiply by 65 and [×] dollars per gram (T: 0, Z: 0, Y: 0, X: 602160), and tap [$] (T: 0, Z: 0, Y: 0, X: 602160.00) to format the ingot's value at $\mathbf{\$602,160.00}$!

Chapter 07: Discrete Probability, Primes & Pre-Algebra — Extended Worked Problems

Accompanies Chapter 07 of From Blocks to the Stack (52 Worked Practice Problems)

Workbook Problem 7.1: Compound Coin Toss: Three Consecutive Heads

Triple Coin Toss Tournament Tiebreaker: A match tiebreaker requires flipping three consecutive heads with a balanced coin ($P = (1/2)^3$). Determine the compound probability of three consecutive heads. Arrange a 3-tier binary tree using two-color counters. Out of 8 possible paths ($\text{HHH}, \text{HHT}, \dots, \text{TTT}$), exactly 1 path yields three consecutive heads.

Cube the single-flip probability ($0.5^3$) on the SC-6: 0.5 [ENTER] [3] [g] [yˣ] $\to$ 0.125 (T: 0, Z: 0, Y: 0, X: 0.125). Every flip cuts your surviving branch in half ($1/2 \to 1/4 \to 1/8$), leaving a $\mathbf{0.1250}$ ($12.5\%$) chance of three heads in a row.

Workbook Problem 7.2: Sum of Two Dice: Rolling an 11 or 12

High-Rollers Two-Dice Roll: Two dice are rolled simultaneously in the final round. The sum is 11 for two combinations ($(5,6), (6,5)$) and 12 for one combination ($(6,6)$). Find the probability of rolling an 11 or 12. Map the $6 \times 6$ grid of 36 equally likely outcomes. Highlight the 3 winning combinations in the top-right corner.

Find the decimal odds for those 3 corner cells out of 36: [3] [ENTER] 36 [÷] $\to$ 0.0833333333 (T: 0, Z: 0, Y: 0, X: 0.0833). Notice why 11 is twice as likely as 12—rolling $(5,6)$ or $(6,5)$ gives you two ways to make 11, while $(6,6)$ is the lone path to 12 ($3/36 = 1/12 \approx \mathbf{8.33\%}$).

Workbook Problem 7.3: Complementary Probability

Complementary Event Probability: The championship venue weather report indicates a $35\%$ probability of rain ($P = 0.35$). Determine the complementary probability of fair weather ($P' = 1 - P$). On the 10-sector probability spinner, $3.5$ sectors represent rain while $6.5$ sectors represent fair weather, illustrating the complement rule $1 - P$.

Subtract the rain chance from the whole certainty of $1$: [1] [ENTER] 0.35 [-] $\to$ 0.65 (T: 0, Z: 0, Y: 0, X: 0.65). Because rain and dry skies partition 100% of the possibilities, the complementary chance of fair weather is $\mathbf{0.6500}$ ($65\%$).

Workbook Problem 7.4: Prime Fortress Barbican Trap Door: Primality of 91

The fortress barbican is guarded by a deceptive trap door set to trigger on composite impostor codes. Evaluate candidate code 91. Take 21 tiles ($7 \times 3$) and arrange them into 3 rows of 7 with zero gaps to demonstrate composite grouping. Now test 91 on your SC-6 against prime factor 7 ($91 \div 7 = 13$), exposing 91 as composite.

Unmask the famous prime impostor 91 on your SC-6: 91 [PRIME] $\to$ 91 (T: 0, Z: 0, Y: 0, X: 91). Even though 91 looks odd and dodges 2, 3, and 5, it is secretly $7 \times 13$ ($70 + 21$)—so the status line flashes COMPOSITE while preserving $\mathbf{91}$ in Register $X$!

Workbook Problem 7.5: Prime Fortress Moat Portcullis: Primality of 143

The iron portcullis over the fortress moat evaluates candidate 143. Determine if the credential is prime or composite. Take 22 tiles ($11 \times 2$) to see how groups of 11 assemble into neat rectangles. For 143, note that $143 \div 11 = 13$, confirming factor pairs beyond 1 and itself.

Test 143 on the SC-6: 143 [PRIME] $\to$ 143 (T: 0, Z: 0, Y: 0, X: 143). Notice the middle digit of 143 is the sum of its outer digits ($1 + 3 = 4$), the classic hallmark of a multiple of 11 ($11 \times 13 = 143$)—flashing COMPOSITE and keeping $\mathbf{143}$ in Register $X$!

Workbook Problem 7.6: Prime Fortress Armory Key: Prime Factorization of 72

The fortress armory lock requires the exact prime decomposition string of tumbler code 72. Decompose 72 into prime factors. Take three Blue 2-blocks ($2 \times 2 \times 2 = 8$) and two Amber 3-blocks ($3 \times 3 = 9$). Stack them vertically on the Prime Factor Studio stage: their combined height ($3 \times 12.64\text{ mm} + 2 \times 20.04\text{ mm} = 78.00\text{ mm}$) aligns squarely with the 72 mark on the comparator wall ($8 \times 9 = 72$).

Reveal the prime building blocks of 72 with [g] [FACT]: 72 [g] [FACT] $\to$ 72 (T: 0, Z: 0, Y: 0, X: 72). The screen displays 2³ × 3², matching your three Blue 2-blocks and two Amber 3-blocks ($8 \times 9 = 72$) to disengage the armory lock!

Workbook Problem 7.7: Prime Fortress Citadel Key: Prime Factorization of 210

The inner citadel vault is sealed with master product code 210. Find its complete prime factor decomposition. Select one 2-block, one 3-block, one 5-block, and one 7-block. Stacking these four distinct prime towers models the square-free product $2 \times 3 \times 5 \times 7 = 210$.

Factor 210 on the SC-6 display: 210 [g] [FACT] $\to$ 210 (T: 0, Z: 0, Y: 0, X: 210). The screen displays 2 × 3 × 5 × 7—the product of the first four prime numbers (called a primorial)!

Workbook Problem 7.8: Prime Fortress High Tower Vault Key: Prime Factorization of 504

The high observatory tower is protected by an astronomical key code of 504. Determine its prime factor components. Factor 504 using prime blocks: three 2-blocks ($2^3 = 8$), two 3-blocks ($3^2 = 9$), and one 7-block ($7$). Product: $8 \times 9 \times 7 = 504$.

Extract the prime factor tree for 504: 504 [g] [FACT] $\to$ 504 (T: 0, Z: 0, Y: 0, X: 504). The display reveals 2³ × 3² × 7, showing that 504 is simply the armory code 72 ($2^3 \times 3^2$) multiplied by one Red 7-block!

Workbook Problem 7.9: Fortress Twin Bastions Gear Synchronization: $\gcd(48, 72)$

Twin drawbridge drive winches with 48 teeth and 72 teeth must mesh via a common drive gear. Find the greatest common factor $\gcd(48, 72)$. Model $48 = 2^4 \times 3$ and $72 = 2^3 \times 3^2$. Place shared prime blocks into the central intersection well: three 2-blocks and one 3-block $\implies 2^3 \times 3 = 24$.

Find the largest shared gear step on the stack: 48 [ENTER] 72 [GCF] $\to$ 24 (T: 0, Z: 0, Y: 0, X: 24). A drive gear with $\mathbf{24\text{ teeth}}$ turns the 48-tooth winch in exactly 2 rotations and the 72-tooth winch in 3 rotations without slipping a single tooth.

Workbook Problem 7.10: Fortress Water Gate Sluice Synchronization: $\gcd(84, 140)$

Two hydraulic sluice floodgates operate on cycles of 84 seconds and 140 seconds. Find the greatest common synchronization interval $\gcd(84, 140)$. Factor $84 = 2^2 \times 3 \times 7$ and $140 = 2^2 \times 5 \times 7$. Shared intersection: two 2-blocks and one 7-block $\implies 2^2 \times 7 = 28$.

Compute $\gcd(84, 140)$ using the [GCF] key: 84 [ENTER] 140 [GCF] $\to$ 28 (T: 0, Z: 0, Y: 0, X: 28). Both floodgate timers divide cleanly into $\mathbf{28\text{-second}}$ beats ($84 = 3 \times 28$ and $140 = 5 \times 28$).

Workbook Problem 7.11: Fortress Sentry Patrol Synchronization: $\operatorname{lcm}(15, 20)$

Two rampart sentry patrols cycle every 15 minutes and 20 minutes respectively. Find the least common multiple $\operatorname{lcm}(15, 20)$ when both sentries meet at the gate. Model $15 = 3 \times 5$ and $20 = 2^2 \times 5$. The union takes maximum powers: two 2-blocks, one 3-block, and one 5-block $\implies 2^2 \times 3 \times 5 = 60$. Stacking blocks aligns with mark 60.

Evaluate the patrol meeting interval with [f] [LCM]: 15 [ENTER] 20 [f] [LCM] $\to$ 60 (T: 0, Z: 0, Y: 0, X: 60). After 4 laps for the 15-minute sentry and 3 laps for the 20-minute sentry, both guards meet at the gate every $\mathbf{60\text{ minutes}}$ ($1\text{ hour}$).

Workbook Problem 7.12: Fortress Signal Beacon Cycle: $\operatorname{lcm}(12, 18)$

Two optical signal beacons on the fortress bastions flash at intervals of 12 seconds and 18 seconds. Find the least common interval $\operatorname{lcm}(12, 18)$ when both flash in unison. Decompose $12 = 2^2 \times 3$ and $18 = 2 \times 3^2$. The union of prime blocks takes two 2-blocks and two 3-blocks $\implies 4 \times 9 = 36$.

Calculate the synchronized flash cycle on the stack: 12 [ENTER] 18 [f] [LCM] $\to$ 36 (T: 0, Z: 0, Y: 0, X: 36). Beacon A flashes 3 times and Beacon B flashes 2 times to light up the night sky together every $\mathbf{36\text{ seconds}}$.

Workbook Problem 7.13: Fortress Supply Depot Allocation: Euclidean Division $148 \div 32$

In the fortress provisions depot, 148 ration packs are organized into standard field cases holding 32 packs each. Find the number of full cases packed and remaining loose packs using [g] [÷R]. Docking 1, 4, and 8 on your Place Value Tray (PV-01) (or modeling a smaller proportional batch like $14 \div 3 = 4\text{ R }2$ with your 24 Counting Tiles first), grouping 148 into sets of 32 fills 4 cases ($4 \times 32 = 128$) and leaves 20 packs remaining.

Compute the quotient and remainder in a single tap: 148 [ENTER] 32 [g] [÷R] $\to$ 4 (T: 0, Z: 0, Y: 20, X: 4). Instead of a messy decimal ($4.625$), Register $X$ gives you 4 full cases while Register $Y$ hands you the 20 leftover packs directly!

Workbook Problem 7.14: Fortress Treasury Safe Cipher Wheel: $\gcd(24, 36)$

The fortress treasury safe wheel requires identifying the greatest shared factor of 24 and 36 to engage the master locking tumblers. List all divisors: for 24: $\{1, 2, 3, 4, 6, 8, 12, 24\}$; for 36: $\{1, 2, 3, 4, 6, 9, 12, 18, 36\}$. The shared common factor set is $\{1, 2, 3, 4, 6, 12\}$, with greatest factor 12.

Confirm the greatest common factor on the SC-6: 24 [ENTER] 36 [GCF] $\to$ 12 (T: 0, Z: 0, Y: 0, X: 12). The largest locking step shared by both safe wheels is 12.

Workbook Problem 7.15: Median of an Even Dataset

Environmental Temperature Monitoring: Six ambient temperatures are recorded: $68, 72, 74, 76, 80, 84^\circ\text{F}$. Find the median temperature (average of two central elements 74 and 76). Line up six Counting Tiles in ascending order. Because $n = 6$ is even, there is no single middle tile; average the middle pair (74 and 76) to find their midpoint: $\frac{74+76}{2} = 75^\circ\text{F}$.

Evaluate the even-sample median on your SC-6: [g] [CLΣ] 68 [Σ+] 72 [Σ+] 74 [Σ+] 76 [Σ+] 80 [Σ+] 84 [Σ+] [f] [MED] $\to$ 75 (T: 0, Z: 0, Y: 6, X: 75). The SC-6 automatically sorts all $n=6$ readings and averages the middle pair ($74$ and $76$) to report $\mathbf{75.00^\circ\text{F}}$.

Workbook Problem 7.16: Mode of a Shoe Size Survey

Footwear Inventory Survey: Shoe sizes collected from 8 customers: $6, 7, 7, 7, 8, 8, 9, 10$. Find the statistical mode (most frequent size). Stack Counting Tiles for each shoe size. Size 7 forms the tallest column with 3 tiles, visually standing out as the statistical mode.

Extract the mode from the statistical registers: [g] [CLΣ] [6] [Σ+] [7] [Σ+] [7] [Σ+] [7] [Σ+] [8] [Σ+] [8] [Σ+] [9] [Σ+] 10 [Σ+] [g] [MODE] $\to$ 7 (T: 0, Z: 0, Y: 8, X: 7). For a shoe store manager, the mode (size 7) is more useful than the mean because you can't stock a fractional shoe size!

Workbook Problem 7.17: Range of High Temperatures

Weather Station Spread Analysis: Weekly maximum temperatures: $62, 65, 78, 81, 74, 69, 85^\circ\text{F}$. Calculate the temperature range ($\max - \min = 85 - 62$). Mark the minimum temperature ($62^\circ$) and maximum temperature ($85^\circ$) on a number line. The span between the two extremes measures $23^\circ$.

Calculate the weekly spread using [RNG]: [g] [CLΣ] 62 [Σ+] 65 [Σ+] 78 [Σ+] 81 [Σ+] 74 [Σ+] 69 [Σ+] 85 [Σ+] [RNG] $\to$ 23 (T: 0, Z: 0, Y: 0, X: 23). Even though the temperatures were entered out of order, the SC-6 finds the coolest ($62^\circ$) and warmest ($85^\circ$) days to report a $\mathbf{23.00^\circ\text{F}}$ range.

Workbook Problem 7.18: Interquartile Range of Daily Rainfall

Precipitation Spread: Weekly rainfall measurements: $0.1, 0.2, 0.4, 0.6, 0.9, 1.2, 1.5\text{ inches}$. Find the interquartile range ($Q_3 - Q_1$). Using the SC-6 exclusive-median quartile method (matching Big Ideas Math), excluding the median ($0.6$) leaves the lower half ($0.1, 0.2, 0.4$) with midpoint $Q_1 = 0.20$ and the upper half ($0.9, 1.2, 1.5$) with midpoint $Q_3 = 1.20$, giving $\text{IQR} = 1.20 - 0.20 = 1.00\text{ inch}$.

Compute the middle-50% spread directly with [IQR]: [g] [CLΣ] 0.1 [Σ+] 0.2 [Σ+] 0.4 [Σ+] 0.6 [Σ+] 0.9 [Σ+] 1.2 [Σ+] 1.5 [Σ+] [IQR] $\to$ 1 (T: 0, Z: 0, Y: 0, X: 1). Unlike the full range, the interquartile range ($\mathbf{1.00\text{ inch}}$) ignores the driest and stormiest outlier days to show how typical rainfall varies.

Workbook Problem 7.19: Mean Absolute Deviation of a 5-Point Dataset

Sensor Error Analysis: A 5-point dataset: $10, 15, 20, 25, 30$ has mean $\bar{x} = 20.0$. Deviations from mean are $|-10|, |-5|, |0|, |+5|, |+10|$. Sum of deviations $= 30$. Find the Mean Absolute Deviation ($\text{MAD} = 30 / 5 = 6.0$). Lay out the points on a number line. Measure the distance from each point to the center mark 20: $10 + 5 + 0 + 5 + 10 = 30$. Dividing by 5 yields the average distance from center.

Find the Mean Absolute Deviation on the SC-6: [g] [CLΣ] 10 [Σ+] 15 [Σ+] 20 [Σ+] 25 [Σ+] 30 [Σ+] [g] [MAD] $\to$ 6 (T: 0, Z: 0, Y: 0, X: 6). On average, each sensor reading sits 6 units away from the balance point of 20.

Workbook Problem 7.20: Frequency of Score 85

Scorecard Frequency Binning: In survey dataset $70, 75, 80, 80, 85, 85, 85, 90, 90, 95$, accumulate observation 85 and inspect active sample counter. Place a Counting Tile into the score 85 column as the calculator logs the entry into the statistical registers.

Log the observation into [Σ+]: 85 [Σ+] $\to$ 1 (T: 0, Z: 0, Y: 0, X: 1). Every time you press [Σ+], Register $X$ shows the updated sample count $n$ so you never lose your place in a long list of numbers.

Workbook Problem 7.21: Subtracting a Negative Number

Evaluate $12 - (-18)$. Try a tabletop example with two-color counters by placing positive yellow counters and introducing zero pairs (1 yellow $+1$ paired with 1 crimson $-1$). Removing the negative crimson counters leaves both sets of positive yellow counters ($12 + 18 = 30$).

Key in $12 - (-18)$ using [+/-] and [-]: 12 [ENTER] 18 [+/-] [-] $\to$ 30 (T: 0, Z: 0, Y: 0, X: 30). Taking away an 18-point penalty has the exact same effect as awarding 18 bonus points ($12 + 18 = \mathbf{30}$).

Workbook Problem 7.22: Multiplying Signed Numbers (Negative times Negative)

Evaluate $(-7) \times (-8)$. On a number line, facing the negative direction and walking backward reverses your orientation $180^\circ$ twice, moving you forward into positive territory to $+56$.

Multiply the two negative factors on the stack: [7] [+/-] [ENTER] [8] [+/-] [×] $\to$ 56 (T: 0, Z: 0, Y: 0, X: 56). Removing 7 debts of $\$8$ each leaves you $\$56$ richer!

Workbook Problem 7.23: Dividing Signed Numbers (Negative divided by Positive)

Evaluate $(-72) \div (+9)$. Partitioning a total debt of $-72$ evenly across 9 equal groups places a debt of $-8$ into each group.

Divide $-72$ by $9$ on the SC-6: 72 [+/-] [ENTER] [9] [÷] $\to$ -8 (T: 0, Z: 0, Y: 0, X: -8). Splitting a 72-meter submarine dive into 9 equal stages means descending $-8\text{ meters}$ per stage.

Workbook Problem 7.24: Absolute Value of Negative Integer

Evaluate $|-125|$. Locate $-125$ on a number line. Regardless of direction from the origin, geometric distance from 0 to $-125$ is strictly non-negative: 125 units.

Measure the distance from zero using [ABS]: 125 [+/-] [ABS] $\to$ 125 (T: 0, Z: 0, Y: 0, X: 125). Whether you walk 125 steps east or 125 steps west, your pedometer logs 125 steps of distance.

Workbook Problem 7.25: Absolute Value Expression: $|-18| + |+12|$

Evaluate $|-18| + |12|$. On a number line, the distance from $-18$ to 0 is 18 units and the distance from 0 to $+12$ is 12 units, giving a combined distance of 30 units.

Add the two magnitudes on the stack: 18 [+/-] [ABS] 12 [ABS] [+] $\to$ 30 (T: 0, Z: 0, Y: 0, X: 30). Because [ABS] acts immediately on Register $X$, you never need vertical-bar brackets around each term to find the total distance of 30 units.

Workbook Problem 7.26: Number Line Distance: Between $-12$ and $+15$

Find distance between $-12$ and $+15$ on the real line: $|15 - (-12)|$. Mark $-12$ and $+15$ on a number line and count the unit intervals between them: 12 units to reach zero, plus 15 units past zero, totaling 27 units.

Compute the span with subtraction and [ABS]: 15 [ENTER] 12 [+/-] [-] [ABS] $\to$ 27 (T: 0, Z: 0, Y: 0, X: 27). Subtracting the two coordinates and tapping [ABS] always gives the positive distance (27 units), no matter which endpoint you enter first!

Workbook Problem 7.27: Square of Negative Number: $(-9)^2$

Evaluate $(-9)^2$. Take 9 tiles from your kit and assemble a $3 \times 3$ grid ($(-3)^2 = +9$) to see that squaring a negative value yields a positive result. Scaling up to $(-9)^2$ produces $+81$.

Square $-9$ in place on the SC-6: [9] [+/-] [x²] $\to$ 81 (T: 0, Z: 0, Y: 0, X: 81). On an algebraic calculator, typing -9² without parentheses gives -81 because it squares 9 before negating; on the SC-6, -9 is already a signed number in Register $X$, so [x²] multiplies $(-9) \times (-9) = \mathbf{81}$ cleanly!

Workbook Problem 7.28: Geometric Square Root: $\sqrt{289}$

A square plaza has area $289\text{ m}^2$. Find side length. Model the square root concept on a small scale with your Counting Tiles: assemble a $4 \times 4$ square using 16 tiles to see that edge length is $\sqrt{16} = 4$. For the $289\text{ m}^2$ plaza, taking the square root extracts the side length directly.

Extract the side length with [f] [√x]: 289 [f] [√x] $\to$ 17 (T: 0, Z: 0, Y: 0, X: 17). Because $17$ is a prime number, a 289-tile plaza can only be arranged as a $\mathbf{17\text{ m}} \times 17\text{ m}$ square (or a $1 \times 289$ strip)!

Workbook Problem 7.29: Geometric Square Root: $\sqrt{625}$

A square chessboard has area $625\text{ cm}^2$. Find side length. Using your Place Value flats and rods, notice that $25 \times 25 = 625$, so each edge of the square board measures 25 cm.

Confirm $\sqrt{625}$ on the SC-6: 625 [f] [√x] $\to$ 25 (T: 0, Z: 0, Y: 0, X: 25). Every edge of the chessboard measures $\mathbf{25\text{ cm}}$ ($5^2\text{ cm}$), since $625 = 5^4$.

Workbook Problem 7.30: Random Number in Range 10 to 50

Generate a random integer between 10 and 50 using [g] [t…x]. Just as spinning the 10-sector Probability Spinner generates uniform random digits, the [g] [t…x] key samples every whole number from the lower bound in Register $Y$ ($10$) to the upper bound in Register $X$ ($50$) with equal likelihood.

Sample the bounded range [10, 50] on the stack: 10 [ENTER] 50 [g] [t…x] $\to$ 28 (T: 0, Z: 0, Y: 10, X: 28). Each of the 41 integers from 10 through 50 has an identical $1/41$ chance of landing in Register $X$.

Workbook Problem 7.31: Divisibility by 9 Check via Remainder

Check if 1,557 is divisible by 9 using Euclidean division ([g] [÷R]). First, try a small example with your Place Value kit: test divisibility by 9 on 18 (1 ten-bar and 8 unit cubes), grouping into 2 sets of 9 with zero remainder ($18 \div 9 = 2\text{ R }0$). Now check 1,557 on the stack without needing 1,557 blocks:

Divide 1,557 by 9 with [g] [÷R]: 1557 [ENTER] [9] [g] [÷R] $\to$ 173 (T: 0, Z: 0, Y: 0, X: 173). Notice why the remainder in Register $Y$ is $0$: the digits of $1,557$ add up to $1 + 5 + 5 + 7 = 18$, which is a multiple of 9 ($173 \times 9 = 1,557$)!

Workbook Problem 7.32: LCD of Three Denominators: $\frac{1}{4}, \frac{1}{6}, \frac{1}{10}$

Model common multiples with your kit tiles by aligning 4-tile and 6-tile rows until their edges meet flush at 12 tiles ($\operatorname{lcm}(4,6)=12$), and then use your SC-6 to include a third denominator of 10: [4] [ENTER] [6] [f] [LCM] 10 [f] [LCM] $\rightarrow$ 60.

This sequence finishes with $(T: 0, Z: 0, Y: 0, X: 60)$, chaining the two LCM operations directly on the stack ($\operatorname{lcm}(4,6) = 12$, then $\operatorname{lcm}(12,10) = 60$) to find the three-way Least Common Denominator of 60.

Workbook Problem 7.33: Temperature Range Across a Sub-Zero Week

Lowest temperature was $-22^\circ\text{F}$, highest was $+14^\circ\text{F}$. Find total temperature swing. On a vertical thermometer scale, mark $-22$ below zero and $+14$ above zero. Trace the rise from $-22$ to 0 (22 degrees) and from 0 to $+14$ (14 degrees), measuring the total thermal swing of 36 degrees.

Subtract $-22$ from $14$ on the stack: 14 [ENTER] 22 [+/-] [-] $\to$ 36 (T: 0, Z: 0, Y: 0, X: 36). Climbing $22^\circ$ up to zero and another $14^\circ$ above zero produces a total weather swing of $\mathbf{36^\circ\text{F}}$.

Workbook Problem 7.34: Rolling Double Sixes with Two Dice

What is the probability of rolling double sixes with two fair six-sided dice ($1$ favorable outcome out of $36$)? Examine a $6 \times 6$ grid of dice pairings (36 cells). Only one cell represents $(6,6)$. Compute the decimal share of that single cell.

Divide 1 by 36 on the SC-6: [1] [ENTER] 36 [÷] $\to$ 0.0277777778 (T: 0, Z: 0, Y: 0, X: 0.0278). Out of all 36 cells in the two-dice matrix, "boxcars" $(6,6)$ occupies just a single corner cell ($\approx \mathbf{2.78\%}$).

Workbook Problem 7.35: Probability of Not Rolling a 6 on a Die

Five faces are not 6. Compute $5/6$. Inspect the 6 faces on a physical die from your kit: $\{1, 2, 3, 4, 5, 6\}$. Excluding the 6-face leaves 5 faces for the complement event ($E'$).

Evaluate $5 \div 6$ on the stack: [5] [ENTER] [6] [÷] $\to$ 0.8333333333 (T: 0, Z: 0, Y: 0, X: 0.8333). Five of the six faces dodge the number 6, giving an $\mathbf{83.33\%}$ chance of rolling 1 through 5.

Workbook Problem 7.36: Prime Factor Decomposition of 1,001

Decompose 1,001 into prime factors (a classic number-theory product). Take 12 unit cubes from your Place Value kit and build a $2 \times 2 \times 3$ prism to see how three prime dimensions multiply into a single 3D volume. Now factor 1,001 on your SC-6 to discover its three prime dimensions $7 \times 11 \times 13$:

Factor 1,001 with [g] [FACT]: 1001 [g] [FACT] $\to$ 1001 (T: 0, Z: 0, Y: 0, X: 1001). Known as the "Arabian Nights number," $1,001$ factors into three consecutive primes: 7 × 11 × 13! (Try multiplying any 3-digit number like $452 \times 1,001 = 452,452$—dividing it by 7, 11, and 13 brings your secret number right back!)

Workbook Problem 7.37: Prime Factor Decomposition of 1,729 (Hardy-Ramanujan Number)

Decompose 1,729 into prime factors. Known as the Hardy-Ramanujan taxicab number, 1,729 is the smallest positive integer expressible as the sum of two cubes in two different ways ($1^3 + 12^3 = 9^3 + 10^3$). Use your SC-6 to reveal its three prime factors:

Decompose 1,729 on the display: 1729 [g] [FACT] $\to$ 1729 (T: 0, Z: 0, Y: 0, X: 1729). When the mathematician G. H. Hardy visited Srinivasa Ramanujan in the hospital and called cab #1729 "a rather dull number," Ramanujan immediately replied that it is the smallest number expressible as the sum of two cubes in two ways—and on your SC-6, [g] [FACT] shows it also splits into three primes spaced 6 units apart: 7 × 13 × 19!

Workbook Problem 7.38: Weighted Mean of Classroom Grades

A grading syllabus weights homework (average 90, weight 2), quizzes (average 82, weight 3), and exams (average 88, weight 5) for a total weight of 10. Accumulate the weighted points directly on the stack: 90 [ENTER] [2] [×] 82 [ENTER] [3] [×] [+] 88 [ENTER] [5] [×] [+] 10 [÷] $\rightarrow$ 86.6.

This sequence finishes with $(T: 0, Z: 0, Y: 0, X: 86.6)$, computing the weighted grade average of $86.6\%$ without scratch paper or order-of-operations ambiguity.

Workbook Problem 7.39: School Book Fair & Robotics Club Treasury Across Four Seasons

The district student council tracks its combined STEM Club & Book Fair treasury across four school seasons: Fall Book Fair brings in $+\$14,500$, Winter 3D-printer filament and kit restocking spends $-\$3,200$, Spring Maker Market earns $+\$8,900$, and Summer camp bus chartering spends $-\$1,400$. Try a small tabletop test with yellow credit and crimson debit counters to see how expenses cancel part of each fundraiser, then combine the four seasonal cash flows on the stack: 14500 [ENTER] 3200 [-] 8900 [+] 1400 [-] $\rightarrow$ 18800.

This sequence finishes with $(T: 0, Z: 0, Y: 0, X: 18800)$, yielding a net year-end student club treasury balance of $\$18,800.00$.

Workbook Problem 7.40: Drawing Consecutive Matching Socks

A drawer holds 6 black socks and 4 blue socks (10 total). The probability of drawing a first black sock is $6/10$, and drawing a second black sock without replacement is $5/9$. Evaluate the dependent pair probability on the stack: [6] [ENTER] 10 [÷] [5] [ENTER] [9] [÷] [×] $\rightarrow$ 0.3333333333.

This sequence finishes with $(T: 0, Z: 0, Y: 0, X: 0.3333333333)$, proving an exact $1/3$ ($33.33\%$) chance of pulling two black socks in the dark.

Workbook Problem 7.41: Prime Factorization of 3,600

Decompose 3,600 into prime factors ($3600 = 60^2$). Constructing a factor tree shows $3600 = 36 \times 100 = 2^4 \times 3^2 \times 5^2$.

Verify the prime factor decomposition on your SC-6: 3600 [g] [FACT] $\to$ 3600 (T: 0, Z: 0, Y: 0, X: 3600). Look at the exponents in 2⁴ × 3² × 5²: every single exponent is an even number ($4, 2, 2$), which is why $3,600$ is a perfect square ($60^2$)—and why there are $3,600$ seconds in an hour!

Workbook Problem 7.42: GCD and LCM of 45 and 75

Decomposing $45 = 3^2 \times 5$ and $75 = 3 \times 5^2$ reveals their prime intersection $3 \times 5 = 15$ ($\gcd$) and prime union $3^2 \times 5^2 = 225$ ($\operatorname{lcm}$). Compute both dual parameters on the stack: 45 [ENTER] 75 [GCF] 45 [ENTER] 75 [f] [LCM] $\rightarrow$ 225.

This sequence finishes with $(T: 0, Z: 0, Y: 15, X: 225)$, displaying $\text{LCM} = 225$ in Register $X$ while preserving $\text{GCF} = 15$ in Register $Y$ ($15 \times 225 = 3,375 = 45 \times 75$).

Workbook Problem 7.43: Synchronized Municipal Traffic Signal Cycles

Signal A cycles every $45\text{ seconds}$, Signal B cycles every $60\text{ seconds}$. Find the synchronized cycle time: $\operatorname{lcm}(45, 60)$. Listing multiples of 45 ($45, 90, 135, 180\dots$) and multiples of 60 ($60, 120, 180\dots$) shows that their first shared multiple is 180 seconds.

Find $\operatorname{lcm}(45, 60)$ using [f] [LCM]: 45 [ENTER] 60 [f] [LCM] $\to$ 180 (T: 0, Z: 0, Y: 0, X: 180). Both traffic lights turn green together every 180 seconds ($3\text{ minutes}$).

Workbook Problem 7.44: Euclidean Division Ladder for $\gcd(546, 210)$

Compute Euclidean steps for $546$ and $210$: $546 \div 210 = 2\text{ R }126$; $210 \div 126 = 1\text{ R }84$; $126 \div 84 = 1\text{ R }42$; $84 \div 42 = 2\text{ R }0$. Each remainder becomes the new divisor for the next step until the remainder reaches zero, leaving 42 as the last non-zero remainder.

Verify the Euclidean ladder endpoint on the SC-6: 546 [ENTER] 210 [GCF] $\to$ 42 (T: 0, Z: 0, Y: 0, X: 42). In one keystroke, the hardware [GCF] engine zips down all four rungs of Euclid's ladder to reveal $\gcd(546, 210) = \mathbf{42}$.

Workbook Problem 7.45: Mean Absolute Deviation of a 7-Element Dataset

Dataset: $12, 14, 15, 16, 18, 20, 24$ (Mean = 17.0). Deviations: $5, 3, 2, 1, 1, 3, 7$. Sum = 22. MAD = $22/7$. Plot the 7 points on a dot plot and mark the mean at 17. Measuring the distance from each dot to 17 gives $5 + 3 + 2 + 1 + 1 + 3 + 7 = 22$, and dividing among the 7 data points gives $22/7 \approx 3.1429$.

Evaluate the Mean Absolute Deviation with [g] [MAD]: [g] [CLΣ] 12 [Σ+] 14 [Σ+] 15 [Σ+] 16 [Σ+] 18 [Σ+] 20 [Σ+] 24 [Σ+] [g] [MAD] $\to$ 3.1428571429 (T: 0, Z: 0, Y: 0, X: 3.1428571429). Look closely at the decimal $3.142857\dots$ in Register $X$: because the sum of deviations is $22$ across $n=7$ points, the MAD is the famous fraction $22/7$ (Archimedes' classic approximation for $\pi$)!

Workbook Problem 7.46: Interquartile Range Across Grade 6 Science Scores

Scores: $65, 72, 75, 80, 84, 88, 92, 96$. Find IQR ($Q_3 - Q_1 = 90 - 73.5 = 16.5$). Line up 8 Counting Tiles to mark the 8 sorted scores, splitting them into a lower half ($65, 72, 75, 80$) with midpoint $Q_1 = 73.5$ and an upper half ($84, 88, 92, 96$) with midpoint $Q_3 = 90$. The spread between the quartiles is $90 - 73.5 = 16.5$.

Compute the interquartile range on the stack: [g] [CLΣ] 65 [Σ+] 72 [Σ+] 75 [Σ+] 80 [Σ+] 84 [Σ+] 88 [Σ+] 92 [Σ+] 96 [Σ+] [IQR] $\to$ 16.5 (T: 0, Z: 0, Y: 0, X: 16.5). Across the $n=8$ students, the middle 50% of the class spans 16.5 points in Register $X$.

Workbook Problem 7.47: Chemical Freezing Point Depression

Pure water freezes at $0^\circ\text{C}$. Adding rock salt lowers the freezing point by $-7.8^\circ\text{C}$. Adding calcium chloride drops it an additional $-6.4^\circ\text{C}$. Find final freezing point. On a vertical thermometer scale, move down $7.8^\circ$ below zero and then $6.4^\circ$ further below zero for a combined drop of $14.2^\circ$, reaching $-14.2^\circ\text{C}$.

Sum the two negative temperature drops on the SC-6: 7.8 [+/-] [ENTER] 6.4 [+/-] [+] $\to$ -14.2 (T: 0, Z: 0, Y: 0, X: -14.2). That is why road crews spread de-icing salts before a winter storm: water won't freeze into ice until the thermometer plunges to $-14.2^\circ\text{C}$!

Workbook Problem 7.48: Product of Three Signed Integers

Evaluate $(-4) \times (-5) \times (-6)$. On a number line, multiplying negative by negative reverses direction twice into positive territory: $(-4) \times (-5) = +20$. Multiplying by a third negative factor reverses direction back into negative territory: $+20 \times (-6) = -120$. Multiply all three signed integers on the stack: [4] [+/-] [ENTER] [5] [+/-] [×] [6] [+/-] [×] $\rightarrow$ -120.

Tap 4 [+/-] [ENTER] 5 [+/-] [×] $\to$ (T: 0, Z: 0, Y: 0, X: 20). The first pair of negative factors flips positive ($+20$). Tap 6 [+/-] [×] $\to$ (T: 0, Z: 0, Y: 0, X: -120). Multiplying by the third negative factor flips the sign back to $-120$—showing why any odd number of negative factors produces a negative product!

Workbook Problem 7.49: Order of Operations with Negative Numbers

Evaluate $(-18) \times 4 - (-36) \div (-9) = -72 - 4 = -76$. Model the two terms separately: Term 1 is 4 groups of $-18$ ($-72$), and Term 2 is $(-36) \div (-9) = +4$. Subtracting $+4$ from $-72$ moves 4 units further into negative territory to $-76$. Evaluate both terms without parentheses on the SC-6: 18 [+/-] [ENTER] [4] [×] 36 [+/-] [ENTER] [9] [+/-] [÷] [-] $\rightarrow$ -76.

Tap 18 [+/-] [ENTER] 4 [×] $\to$ (T: 0, Z: 0, Y: 0, X: -72) to lock the first term ($-72$) on the stack. Tap 36 [+/-] [ENTER] 9 [+/-] [÷] $\to$ (T: 0, Z: 0, Y: -72, X: 4) so the second quotient ($+4$) sits in $X$ while $-72$ waits safely in $Y$. Tap [-] $\to$ (T: 0, Z: 0, Y: 0, X: -76) to subtract $-72 - 4 = \mathbf{-76}$ with zero parentheses!

Workbook Problem 7.50: Geometric Square Root of a Square Ranch

A square ranch has an area of $4,096\text{ m}^2$. Find side length. Because $60^2 = 3,600$ and $70^2 = 4,900$, the side length of the square ranch must lie between 60 m and 70 m, with $64 \times 64 = 4,096$.

Find the exact side length using [f] [√x]: 4096 [f] [√x] $\to$ 64 (T: 0, Z: 0, Y: 0, X: 64). Fun computer-science bonus: $4,096$ is $2^{12}$, so its square root is $2^6 = \mathbf{64\text{ m}}$ per side!

Workbook Problem 7.51: Divisibility by 11 Verification

Test if 2,596 is divisible by 11 using Euclidean division ([g] [÷R]). Apply the alternating digit sum divisibility rule for 11: $(2 + 9) - (5 + 6) = 11 - 11 = 0$. Since the alternating difference is a multiple of 11, the number is divisible with zero remainder.

Confirm exact divisibility on the stack: 2596 [ENTER] 11 [g] [÷R] $\to$ 236 (T: 0, Z: 0, Y: 0, X: 236). Register $X$ shows the exact quotient 236 and Register $Y$ shows remainder 0, proving that the alternating-digit trick ($2 - 5 + 9 - 6 = 0$) spotted a true multiple of 11!

Workbook Problem 7.52: Mersenne Number Primality Check: $2^5 - 1 = 31$

Compute $2^5 - 1$ and check if the result is prime. Just as 7 or 13 tiles from your Counting Tile kit can only form a single $1 \times p$ row, evaluating $2^5 - 1 = 31$ on your SC-6 confirms that 31 has no rectangular factors.

Compute $2^5 - 1$ and test primality on the SC-6: [2] [ENTER] [5] [g] [yˣ] [1] [-] [PRIME] $\to$ Flashes (T: 0, Z: 0, Y: 0, X: 31). Named after the French monk Marin Mersenne, primes of the form $2^p - 1$ hold the world record for the largest known prime numbers ever discovered—and $M_5 = 2^5 - 1 = \mathbf{31}$ flashes PRIME right in your hand!

Chapter 08: The Gamified Math Quiz Engine — Extended Worked Problems

Accompanies Chapter 08 of From Blocks to the Stack (29 Worked Practice Problems)

Workbook Drill 8.1 (Level 1): $6 + 3 = ?$

Use your Place Value Tray (PLACE-01). Start with the larger benchmark group of 6 (a full row of 5 plus 1 dot), then count on 3 quick hops: 7, 8, 9 ($6 + 3 = 9$). When 6 3 + = ? appears in Quiz Mode, key in 9 [≠] in under $< 2.0\text{ seconds}$.

On the stack, 6 [ENTER] 3 [+] $\to$ 9 (T: 0, Z: 0, Y: 0, X: 9). Watch 6 lift to Floor $Y$ on [ENTER] before [+] drops it onto 3 to land on 9.

Workbook Drill 8.2 (Level 1): $8 - 0 = ?$

In your Place Value Tray (PLACE-01), set out 8 counters (5 on top, 3 on bottom). Subtracting zero means taking away zero blocks—all 8 counters stay locked in position ($8 - 0 = 8$). For 8 0 - = ? in Quiz Mode, press 8 [≠] on the keypad (target time: $< 1.0\text{ seconds}$).

Testing 8 [ENTER] 0 [-] $\to$ 8 (T: 0, Z: 0, Y: 0, X: 8) shows that subtracting 0 in Register $X$ leaves 8 completely unchanged!

Workbook Drill 8.3 (Level 1): $3 + 7 = ?$

Look at your Place Value Tray (PLACE-01). Place 3 blue counters in the first row. The 7 empty pockets show its exact 10-partner—together they fill all 10 pockets of the tray ($3 + 7 = 10$). When 3 7 + = ? (or Missing-Value Puzzle 3 ? + = 10) shows up in Quiz Mode, tap 10 [≠] (aim for under $< 1.5\text{ seconds}$).

Stack verification: 3 [ENTER] 7 [+] $\to$ 10 (T: 0, Z: 0, Y: 0, X: 10). Just like Textbook Drill 2 in reverse, 3 and 7 snap together on the stack to form a complete 10.

Workbook Drill 8.4 (Level 1): $9 - 5 = ?$

With your Place Value Tray (PLACE-01), a group of 9 fills the top row of 5 plus 4 on the bottom row. When you lift off the full top row of 5, the 4 bottom counters remain ($9 - 5 = 4$). For the Quiz Mode prompt 9 5 - = ?, enter 4 [≠] within $< 1.8\text{ seconds}$.

Keying 9 [ENTER] 5 [-] $\to$ 4 (T: 0, Z: 0, Y: 0, X: 4) takes the benchmark 5 away from 9 in Register $Y$ and drops the difference 4 into Register $X$.

Workbook Drill 8.5 (Level 1): $4 + 4 = ?$

Place matching [4] digit blocks in Row 1 and Row 0 of your Place Value Tray (PLACE-01). Both halves show identical $2 \times 2$ square clusters ($4 + 4 = 8$); doubling 4 clicks together like twin booster rockets! When you see 4 4 + = ? in Quiz Mode, key in 8 [≠] in under $< 1.0\text{ seconds}$.

On your SC-6, 4 [ENTER] 4 [+] $\to$ 8 (T: 0, Z: 0, Y: 0, X: 8). Pro Tip: Because [ENTER] copies 4 into both $Y$ and $X$, you can even test 4 [ENTER] [+] to double a number in three taps!

Workbook Drill 8.6 (Level 1): $10 - 2 = ?$

Start with a full 10 in your Place Value Tray (PLACE-01). Pop out the 2 end counters from the bottom row: you see a full row of 5 plus 3 on the bottom, giving 8 counters ($10 - 2 = 8$). For 10 2 - = ? (or Missing-Value Puzzle 10 ? - = 8) in Quiz Mode, press 8 [≠] (target time: $< 1.2\text{ seconds}$).

Stack trace: 10 [ENTER] 2 [-] $\to$ 8 (T: 0, Z: 0, Y: 0, X: 8). Counting back 2 from 10 on the stack lands cleanly on 8.

Workbook Drill 8.7 (Level 2): $34 + 12 = ?$

Use your two Place Value Tray (PLACE-01): in the left tray, model the tens ($3 + 1 = 4\text{ tens} = 40$); in the right tray, model the ones ($4 + 2 = 6\text{ ones}$). Put them together for $46$. When 34 12 + = ? is shown in Quiz Mode, key in 46 [≠] in under $< 2.5\text{ seconds}$.

Checking 34 [ENTER] 12 [+] $\to$ 46 (T: 0, Z: 0, Y: 0, X: 46) shows how both tens and ones add without regrouping, landing on 46 in Register $X$.

Workbook Drill 8.8 (Level 2): $67 - 25 = ?$

Use your two Place Value Tray (PLACE-01) as a Tens Tray and a Ones Tray: subtract the tens ($6 - 2 = 4\text{ tens} = 40$) and subtract the ones ($7 - 5 = 2\text{ ones}$) to get $42$. On 67 25 - = ? in Quiz Mode, press 42 [≠] (target time: $< 2.8\text{ seconds}$).

In [NUM] mode, 67 [ENTER] 25 [-] $\to$ 42 (T: 0, Z: 0, Y: 0, X: 42). Keeping 67 in Register $Y$ and subtracting 25 in Register $X$ confirms 42.

Workbook Drill 8.9 (Level 2): $47 + 38 = ?$

Use the Make-a-Round-Number Trick with your Place Value Tray (PLACE-01)! Notice that 38 is just 2 shy of 40 (a nearly full 10-group with 8 counters). Add 40 to 47 first ($47 + 40 = 87$), then step back 2 counters to land on $85$. For 47 38 + = ? in Quiz Mode, tap 85 [≠] (aim for under $< 3.5\text{ seconds}$).

Stack check: 47 [ENTER] 38 [+] $\to$ 85 (T: 0, Z: 0, Y: 0, X: 85). Or see how the ones regroup: $7 + 8 = 15$ (1 ten and 5 ones) plus $40 + 30 = 70$ gives 85.

Workbook Drill 8.10 (Level 2): $72 - 45 = ?$

Count Up to Subtract! Use your Place Value Tray (PLACE-01) to bridge the gap from 45 to 72: add 5 counters to 45 to reach the friendly multiple of ten ($50$), then add 22 more ($2\text{ tens} + 2\text{ ones}$) to reach $72$. Your total jump is $5 + 22 = 27$. When 72 45 - = ? (or Missing-Value Puzzle 72 ? - = 27) appears in Quiz Mode, enter 27 [≠] within $< 3.5\text{ seconds}$.

Keying 72 [ENTER] 45 [-] $\to$ 27 (T: 0, Z: 0, Y: 0, X: 27) confirms the difference—counting up from the smaller number to the larger number turns borrowing into easy addition!

Workbook Drill 8.11 (Level 2): $56 + 19 = ?$

Model 19 on your Place Value Tray (PLACE-01)—it fills all 20 pockets except 1. Add 20 to 56 first ($56 + 20 = 76$), then take back that 1 empty pocket ($76 - 1 = 75$). For 56 19 + = ? in Quiz Mode, key in 75 [≠] in under $< 2.8\text{ seconds}$.

Verify on the stack: 56 [ENTER] 19 [+] $\to$ 75 (T: 0, Z: 0, Y: 0, X: 75). Anytime you add 19, 29, or 39, jump up by the next ten and step back 1.

Workbook Drill 8.12 (Level 2): $91 - 38 = ?$

Use Overshoot & Give Back! Since 38 is 2 less than 40, subtract 40 from 91 first ($91 - 40 = 51$). Because you took away 2 too many, pop 2 counters back into your Place Value Tray (PLACE-01) to finish at $51 + 2 = 53$. On 91 38 - = ? in Quiz Mode, press 53 [≠] (target time: $< 3.5\text{ seconds}$).

Stack verification: 91 [ENTER] 38 [-] $\to$ 53 (T: 0, Z: 0, Y: 0, X: 53). Overshooting by 40 and giving back 2 beats crossing out digits every time.

Workbook Drill 8.13 (Level 3): $6 \times 9 = ?$

Use your Place Value Tray (PLACE-01) to see the "10 Minus 1" Rule! Nine is $10 - 1$, so 6 groups of 9 is 6 groups of 10 ($60$) minus 6 single counters ($60 - 6 = 54$). For 6 9 × = ? in Quiz Mode, key in 54 [≠] in under $< 1.5\text{ seconds}$.

Stack trace: 6 [ENTER] 9 [×] $\to$ 54 (T: 0, Z: 0, Y: 0, X: 54). Notice that in 54, the tens digit is $6 - 1 = 5$ and the digits add up to $5 + 4 = 9$.

Workbook Drill 8.14 (Level 3): $48 \div 6 = ?$

Model Half with Your Kit $\to$ Double It! Take all 24 Array Tiles from your kit and arrange them into 6 equal rows: you get 6 rows of 4 ($24 \div 6 = 4$). Since 48 is double 24, double the row length from 4 to 8 ($48 \div 6 = 8$). On 48 6 ÷ = ? (or Missing-Value Puzzle 48 ? ÷ = 8) in Quiz Mode, press 8 [≠] (target time: $< 1.8\text{ seconds}$).

Keying 48 [ENTER] 6 [÷] $\to$ 8 (T: 0, Z: 0, Y: 0, X: 8) confirms 8. Halving 48 to 24 turns a big division fact into one you can build right on your desk.

Workbook Drill 8.15 (Level 3): $12 \times 7 = ?$

Split 12 into $10 + 2$. Use 14 of your 24 Array Tiles to build the $2 \times 7 = 14$ rectangle on your desk, then add the easy $10 \times 7 = 70$ mental block: $70 + 14 = 84$. When 12 7 × = ? flashes in Quiz Mode, tap 84 [≠] (aim for under $< 2.2\text{ seconds}$).

On the stack, 12 [ENTER] 7 [×] $\to$ 84 (T: 0, Z: 0, Y: 0, X: 84). Every $12 \times N$ fact is just $10N + 2N$!

Workbook Drill 8.16 (Level 3): $9 \times 9 = ?$

Use your Place Value Tray (PLACE-01) to picture 9 groups of 10 ($90$) and pull off 1 counter from each of the 9 groups ($90 - 9 = 81$). For 9 9 × = ? in Quiz Mode, enter 81 [≠] within $< 1.2\text{ seconds}$.

Testing 9 [ENTER] 9 [×] $\to$ 81 (T: 0, Z: 0, Y: 0, X: 81) confirms 81—or press 9 [x²] in [NUM] mode to square 9 in a single operation!

Workbook Drill 8.17 (Level 3): $84 \div 12 = ?$

Take all 24 Array Tiles and group them into dozens ($24 \div 12 = 2\text{ dozens}$). Now scale up to 84: since $70 + 14 = 84$, it takes 7 dozens to make 84 ($84 \div 12 = 7$). When 84 12 ÷ = ? (or Missing-Value Puzzle 84 ? ÷ = 7) appears in Quiz Mode, key in 7 [≠] in under $< 2.0\text{ seconds}$.

Stack check: 84 [ENTER] 12 [÷] $\to$ 7 (T: 0, Z: 0, Y: 0, X: 7). Recognizing 84 from Workbook Drill 8.15 ($12 \times 7 = 84$) gives you the inverse division fact for free!

Workbook Drill 8.18 (Level 3): $11 \times 11 = ?$

Use the 11s Split Trick! Model $2 \times 11 = 22$ with 22 of your 24 Array Tiles. For $11 \times 11$, split the digits 1 and 1 apart and put their sum ($1 + 1 = 2$) right in the middle: 121. For 11 11 × = ? in Quiz Mode, press 121 [≠] (target time: $< 1.5\text{ seconds}$).

In [NUM] mode, 11 [ENTER] 11 [×] $\to$ 121 (T: 0, Z: 0, Y: 0, X: 121). $10 \times 11 = 110$, plus one more $11$ equals 121.

Workbook Drill 8.19 (Level 4): $3/4 - 2/5 = ?$

Compare three $1/4$ pieces ($3/4$) against two $1/5$ pieces ($2/5$) in your Fraction Towers comparator well, then find their shared common denominator of 20: $3/4 = 15/20$ and $2/5 = 8/20$, so the height difference is $15/20 - 8/20 = 7/20$. When 3/4 2/5 - = ? appears in Quiz Mode, tap [.] 7 [.] 20 [≠] (aim for under $< 4.5\text{ seconds}$).

In [NUM] mode, . 3 . 4 [ENTER] . 2 . 5 [-] $\to$ 7/20 (T: 0, Z: 0, Y: 0, X: 7/20). Two-Dot entry ([.] 3 [.] 4 and [.] 2 [.] 5) automatically finds the common denominator of 20 on the stack!

Workbook Drill 8.20 (Level 4): $38 \div 6 \to Q=?, R=?$

Take 14 of your 24 Array Tiles and make groups of 6 (2 full groups of 6 with 2 left over). Now scale up to 38: 6 full groups of 6 use $36$ tiles and leave $2$ tiles left over ($Q=6, R=2$). For 38 6 ÷R = ? in Quiz Mode, key in 6 [≠] in under $< 3.0\text{ seconds}$.

On the stack, 38 [ENTER] 6 [g] [÷R] $\to$ 6 (T: 0, Z: 0, Y: 2, X: 6). [g] [÷R] puts the whole quotient 6 in Register $X$ and the remainder 2 right above it in Register $Y$!

Workbook Drill 8.21 (Level 5): $0.25 \times 80 = ?$

Connect decimals to your Fraction Towers! The decimal $0.25$ is stamped right on the $1/4$ tower ($0.25 = 1/4$). Taking one fourth of 80 is just $80 \div 4 = 20$. For 0.25 80 × = ? in Quiz Mode, enter 20 [≠] within $< 2.0\text{ seconds}$.

On the stack, 0.25 [ENTER] 80 [×] $\to$ 20 (T: 0, Z: 0, Y: 0, X: 20). Multiplying by 0.25 is always the same as dividing by 4.

Workbook Drill 8.22 (Level 5): $0.5 \times 48 = ?$

Check the $1/2$ piece in your Fraction Towers kit: $0.5$ is one half! Splitting 48 in half ($48 \div 2 = 24$) leaves exactly 24—the full count of your 24 Array Tiles! For 0.5 48 × = ? in Quiz Mode, enter 24 [≠] within $< 1.5\text{ seconds}$.

Keying 0.5 [ENTER] 48 [×] $\to$ 24 (T: 0, Z: 0, Y: 0, X: 24) confirms 24. Multiplying any even number by 0.5 is as fast as cutting both digits in half ($4 \to 2$, $8 \to 4 \implies 24$).

Workbook Drill 8.23 (Level 5): $10\%\text{ of } 250 = ?$

Use your Place Value Tray (PLACE-01) to see why $10\%$ means 1 counter out of 10 ($\frac{1}{10}$). To find one tenth of 250, simply divide by 10 (shift the decimal point one place left): $250 \div 10 = 25$. On 250 10 % = ? in Quiz Mode, key in 25 [≠] in under $< 1.5\text{ seconds}$.

Stack check: 250 [ENTER] 10 [%] $\to$ 25 (T: 0, Z: 0, Y: 250, X: 25). Notice how [%] keeps the base 250 in Register $Y$ while placing $10\%$ (25) in Register $X$!

Workbook Drill 8.24 (Level 6): $35^2 = ?$

For any number ending in 5, multiply the tens digit $3$ by the next whole number $4$ using 12 of your 24 Array Tiles ($3 \times 4 = 12\text{ hundreds} = 1,200$), then snap on $5^2 = 25$ at the end: $1,225$. When 35 x² = ? flashes in Quiz Mode, tap 1225 [≠] (aim for under $< 1.5\text{ seconds}$).

In [NUM] mode, 35 [x²] $\to$ 1225 (T: 0, Z: 0, Y: 0, X: 1225). The unary [x²] key squares 35 right inside Register $X$ without even needing [ENTER]!

Workbook Drill 8.25 (Level 6): $43 \times 11 = ?$

Use the 11s Middle-Sum Shortcut! Model $4 + 3 = 7$ in your Place Value Tray (PLACE-01). Now spread the digits 4 and 3 apart and drop their sum 7 right into the middle tens place: 473. For 43 11 × = ? in Quiz Mode, enter 473 [≠] within $< 1.5\text{ seconds}$.

On the stack, 43 [ENTER] 11 [×] $\to$ 473 (T: 0, Z: 0, Y: 0, X: 473). Why does it work? Because $43 \times 10 = 430$, and $430 + 43 = 473$.

Workbook Drill 8.26 (Level 6): $65^2 = ?$

Use the Ends-in-5 Squaring Rule! Multiply the tens digit $6$ by the next integer $7$ ($6 \times 7 = 42\text{ hundreds} = 4,200$), then attach $5^2 = 25$ to the last two digits: $4,225$. On 65 x² = ? in Quiz Mode, key in 4225 [≠] in under $< 1.5\text{ seconds}$.

Checking 65 [x²] $\to$ 4225 (T: 0, Z: 0, Y: 0, X: 4225) confirms 4225—once you know $6 \times 7 = 42$, you can type 4225 faster than anyone using an algebraic calculator!

Workbook Drill 8.27 (Level 6): $75^2 = ?$

Recall $7 \times 8 = 56$ from Textbook Drill 5. Multiply the tens digit $7$ by the next integer $8$ ($56\text{ hundreds} = 5,600$) and tack on $25$: $5,625$. For 75 x² = ? in Quiz Mode, press 5625 [≠] (target time: $< 1.5\text{ seconds}$).

Keying 75 [x²] $\to$ 5625 (T: 0, Z: 0, Y: 0, X: 5625) confirms your mental square 5625 in Register $X$ with a single tap of [x²].

Workbook Drill 8.28 (Level 6): $67 \times 11 = ?$

Use the 11s Shortcut with a Carry! Model $6 + 7 = 13$ across your two Place Value Tray (PLACE-01) (1 full ten and 3 ones). Put the 3 in the middle tens slot and carry the 1 ten-group over to the hundreds digit ($6 + 1 = 7$), giving 737. When 67 11 × = ? appears in Quiz Mode, tap 737 [≠] (aim for under $< 1.8\text{ seconds}$).

Stack verification: 67 [ENTER] 11 [×] $\to$ 737 (T: 0, Z: 0, Y: 0, X: 737). Whenever the two digits add to 10 or more, carry 1 to the left digit!

Workbook Drill 8.29 (Level 6): $15\%\text{ of } 80 = ?$

Split $15\%$ into $10\% + 5\%$ using your Place Value Tray (PLACE-01)! $10\%$ of 80 is $8$ counters, and $5\%$ is half of that ($4$ counters). Add them together in your tray: $8 + 4 = 12$. For 80 15 % = ? in Quiz Mode, enter 12 [≠] within $< 1.8\text{ seconds}$.

Testing 80 [ENTER] 15 [%] $\to$ 12 (T: 0, Z: 0, Y: 80, X: 12) shows 80 waiting in Register $Y$ and 12 in Register $X$—so pressing [+] next would even give you $80 + 15\% = 92$.

Chapter 09: Applied Elementary STEM & Maker Labs — Extended Worked Problems

Accompanies Chapter 09 of From Blocks to the Stack (26 Worked Practice Problems)

Workbook Challenge 9.1: Kitchen Model Mass from Density ($\rho = 1.24\text{ g/cm}^3$)

With the funnel volume established at $28.86\text{ cm}^3$, you must convert that cubic volume into mass in grams so you can verify the finished print on your kitchen gram scale. Placing a $1\text{ cm} \times 1\text{ cm} \times 1\text{ cm}$ solid PLA calibration cube on your digital balance yields $1.24\text{ grams}$, establishing the density $\rho = 1.24\text{ g/cm}^3$.

Calculate the total part mass using 28.86 [ENTER] 1.24 [×] $\to$ 35.79 g (T: 0, Z: 0, Y: 0, X: 35.7864). Multiplying volume by polymer density ($m = V \times \rho$) shows that the finished bioplastic funnel will weigh approximately $\mathbf{35.79\text{ grams}}$.

Workbook Challenge 9.2: Spool Material Cost for Kitchen Print ($\$0.02/\text{g}$)

Your maker space purchased a $1,000\text{ gram}$ spool of corn-based PLA for $\$20.00$, establishing an exact material cost ratio of $\$0.02$ per gram. Setting two shiny copper pennies next to each $1\text{ gram}$ calibration weight on your scale pan provides a concrete visual of this unit cost before committing the print job.

Compute the fabrication expense with 35.79 [ENTER] .02 [×] [$] $\to$ $0.72 (T: 0, Z: 0, Y: 0, X: 0.7158). Formatting the result in two-decimal currency mode confirms that fabricating the custom spice funnel costs just $\mathbf{\$0.72}$ in raw bioplastic.

Workbook Challenge 9.3: High-Intensity Green Inspection LED on 9V ($V_f = 2.2\text{V}, I_f = 20\text{mA}$)

To inspect clarity in homemade kombucha fermentation jars, you assemble a high-intensity green flashlight powered by a $9\text{V}$ alkaline battery. Snapping a heavy-duty battery clip onto the terminal and probing with multimeter leads confirms a $9.0\text{ V}$ open-circuit potential, while the green LED specifies a forward drop of $2.2\text{ V}$ at an operating current of $20\text{ mA}$ ($0.020\text{ A}$).

Determine the required current-limiting resistor with [9] [ENTER] 2.2 [-] .020 [÷] $\to$ 340.0 Ω (T: 0, Z: 0, Y: 0, X: 340). Subtracting the forward drop isolates the $6.8\text{ V}$ resistor voltage headroom, and dividing by current yields an exact required resistance of $\mathbf{340.0\,\Omega}$.

Workbook Challenge 9.4: Seed-Bomber Catapult Mean Launch Distance ($150, 162, 156\text{ cm}$)

To reseed a barren community garden plot, you deploy an elastic lever catapult that launches clay wildflower seed balls. Staking colored survey flags along the playground lawn marks landing coordinates across three test launches at $150\text{ cm}$, $162\text{ cm}$, and $156\text{ cm}$.

Accumulate the launches into hardware statistical registers with [g] [CLΣ] 150 [Σ+] 162 [Σ+] 156 [Σ+] [g] [x̄] $\to$ 156.00 cm (T: 0, Z: 0, Y: 0, X: 156). Tapping the mean key recalls the average launch distance of exactly $\mathbf{156.00\text{ cm}}$ across the three trials.

Workbook Challenge 9.5: Catapult Consistency & Mean Absolute Deviation (MAD)

Field accuracy requires understanding launch dispersion around the target line. By placing counting tiles along a tape measure, you find the distances from each landing flag to the $156\text{ cm}$ center: $|150-156| = 6\text{ cm}$, $|162-156| = 6\text{ cm}$, and $|156-156| = 0\text{ cm}$.

Evaluate catapult consistency using [g] [CLΣ] 150 [Σ+] 162 [Σ+] 156 [Σ+] [g] [MAD] $\to$ 4.00 cm (T: 0, Z: 0, Y: 3, X: 4). Averaging the deviations demonstrates high mechanical consistency, with a Mean Absolute Deviation of only $\mathbf{4.00\text{ cm}}$ around the center target.

Workbook Challenge 9.6: Solar Nursery Bean Sprout Growth Velocity ($15\text{ mm} \to 75\text{ mm}$ in $5\text{ days}$)

Inside the solar heated greenhouse nursery, tender pole bean seedlings are monitored daily. Aligning a clear millimeter ruler against the main stem confirms that a test sprout grew from an initial height of $15\text{ mm}$ to $75\text{ mm}$ over a five-day growth trial.

Compute the daily growth rate using 75 [ENTER] 15 [-] [5] [÷] $\to$ 12.0 mm/day (T: 0, Z: 0, Y: 0, X: 12). The stack subtracts baseline height to reveal $60\text{ mm}$ of total expansion, and dividing by 5 days yields an average vertical growth velocity of $\mathbf{12.0\text{ mm/day}}$.

Workbook Challenge 9.7: Total Seedling Height Percentage Surge ($15\text{ mm} \to 75\text{ mm}$)

To assess vegetative vigor under passive solar heating, you calculate the overall percentage surge in seedling height. Modeling a $1:5$ ratio with your 24-tile Counting Studio kit—placing a 3-tile stack ($3 \times 7.0\text{ mm} = 21.0\text{ mm}$ visible $1\text{U}$ height) beside a 15-tile stack ($15 \times 7.0\text{ mm} = 105.0\text{ mm}$ visible $1\text{U}$ height)—provides an immediate physical proof that the new height is five times the starting height, or a $4\times$ ($+400\%$) gain over baseline.

Determine the percentage expansion with 15 [ENTER] 75 [g] [%CHG] $\to$ +400.0% (T: 0, Z: 0, Y: 15, X: 400). The single-tap percent change evaluation shows that the seedling surged by $\mathbf{+400.0\%}$ over its initial height.

Workbook Challenge 9.8: Field Bakery Scaling: Flour for 12 Loaves ($450\text{ g}$ per loaf)

Your solar research expedition needs hearty fresh sourdough bread for a weekend field encampment. A single artisan loaf formulation requires $450\text{ grams}$ of stone-ground flour; you need to scale the recipe to bake a 12-loaf batch for the entire team. Scooping flour into a large mixing tub on a tare-zeroed scale connects bulk pantry purchasing directly to recipe scaling.

Compute the bulk flour requirement with 450 [ENTER] 12 [×] $\to$ 5,400 grams (T: 0, Z: 0, Y: 0, X: 5400). Multiplying loaf flour by batch size yields exactly $\mathbf{5,400\text{ grams}}$ ($5.4\text{ kg}$) of flour for the field bake.

Workbook Challenge 9.9: Artisan Hydration: Water at $65\%$ for $5,400\text{ g}$ Flour

Professional bakers express all ingredient quantities as a percentage of total flour weight (baker's percentages). To achieve an open, tender crumb structure without a sticky dough, the sourdough recipe specifies $65\%$ hydration water. Pouring water into a tared measuring pitcher on your digital gram scale lets you match the exact water weight.

Calculate hydration water using 5400 [ENTER] 65 [%] $\to$ 3,510 grams (T: 0, Z: 0, Y: 5400, X: 3510). Tapping [%] calculates $65\%$ of $5,400$ while preserving the flour base in Register $Y$, confirming you need $\mathbf{3,510\text{ grams}}$ ($3.51\text{ liters}$) of water.

Workbook Challenge 9.10: Baker's Salt Percentage at $2\%$ for $5,400\text{ g}$ Flour

In artisan bread chemistry, fine salt tightens gluten proteins and moderates wild yeast activity, standardizing at exactly $2.0\%$ of flour mass. Weighing out salt crystals into a small pinch bowl on a sensitive gram scale provides a concrete lesson in precision chemical formulation.

Find the salt mass with 5400 [ENTER] [2] [%] $\to$ 108 grams (T: 0, Z: 0, Y: 5400, X: 108). Register $X$ confirms you need $\mathbf{108\text{ grams}}$ of salt, while Register $Y$ continues to hold the $5,400\text{ g}$ flour base ready for further recipe scaling.

Workbook Challenge 9.11: High-Energy Trail Cookies: Scaling $2\ \frac{1}{4}$ Cups Sugar $\times 4$

To supply high-energy snacks for an uphill ridge hike, you quadruple an energy cookie recipe. The base recipe requires $2\ \frac{1}{4}$ cups of dark brown sugar. Stacking four sets of nesting dry measuring cups (two 1-cup scoops and one 1/4-cup scoop each) visually demonstrates the fourfold scaling factor.

Execute mixed-fraction multiplication on the stack: 2 [.] 1 [.] 4 [ENTER] [4] [×] [a ᵇ/c] $\to$ 9 cups (T: 0, Z: 0, Y: 0, X: 9). The SC-6 evaluates the mixed numeral and displays the scaled batch amount as exactly $\mathbf{9\text{ cups}}$ of brown sugar.

Workbook Challenge 9.12: Direct-Drive Extruder Gear Reduction ($16\text{T} \to 64\text{T}$)

To push stiff specialty filaments through a 3D printer hotend without stripping the plastic, you 3D-print a custom gear reduction for the direct-drive extruder. The small motor pinion gear has 16 teeth and meshes into a large 64-tooth driven gear. Marking tooth #1 with a silver felt-tip pen reveals that the motor must turn 4 complete revolutions to turn the large gear once.

Calculate the mechanical reduction ratio: 64 [ENTER] 16 [÷] $\to$ 4.0 : 1 (T: 0, Z: 0, Y: 0, X: 4). Dividing driven teeth by driver teeth delivers an exact $\mathbf{4.0 : 1}$ gear reduction ratio, trading rotational speed for filament-gripping torque.

Workbook Challenge 9.13: Stepper Motor Output RPM ($1,200\text{ RPM} \div 4$)

The NEMA-17 stepper motor driving the extruder spins at an operational speed of $1,200\text{ RPM}$. With the $4:1$ mechanical reduction from Workbook Problem 9.12 installed, you must find the final rotational speed of the hobbed filament feed pulley to program the slicer's microstepping firmware.

Evaluate final shaft speed with 1200 [ENTER] [4] [÷] $\to$ 300 RPM (T: 0, Z: 0, Y: 0, X: 300). Dividing motor speed by the reduction ratio reveals that the filament drive wheel turns smoothly at $\mathbf{300\text{ RPM}}$.

Workbook Challenge 9.14: Hydraulic Syringe Mechanical Advantage ($d_2 = 4\text{ cm}, d_1 = 1\text{ cm}$)

To evaluate the crush resistance of 3D-printed hollow structures, you construct a hydraulic test press using two fluid-filled syringes linked by vinyl tubing. The small input syringe has an internal diameter of $d_1 = 1.0\text{ cm}$, while the heavy output cylinder has a diameter of $d_2 = 4.0\text{ cm}$. By Pascal's Principle, hydraulic force scales with the cross-sectional area ratio: $A_2 / A_1 = (d_2 / d_1)^2$.

Evaluate the hydraulic force multiplier with [4] [ENTER] [1] [÷] [x²] $\to$ 16.0 × (T: 0, Z: 0, Y: 0, X: 16). Squaring the diameter ratio reveals that the hydraulic press delivers a remarkable $\mathbf{16.0\times}$ force multiplication.

Workbook Challenge 9.15: Hydraulic Press Lift Capacity ($10\text{ N input} \times 16$)

With the $16\times$ hydraulic multiplier calibrated, calculate the crushing force exerted on a 3D-printed test bracket when an operator applies an effort force of $10\text{ Newtons}$ (approximately $1\text{ kg}$ of thumb pressure) to the small plunger. Hooking a spring scale to the input plunger confirms the $10\text{ N}$ applied effort.

Calculate the output crush force with 10 [ENTER] 16 [×] $\to$ 160 Newtons (T: 0, Z: 0, Y: 0, X: 160). Multiplying input effort by the mechanical advantage confirms an output force of $\mathbf{160\text{ Newtons}}$, capable of lifting a $16\text{ kg}$ test mass.

Workbook Challenge 9.16: Drone Flight Duration in Minutes and Seconds

To configure the countdown timer on the drone's radio transmitter, convert the decimal flight duration of $11.52\text{ minutes}$ into whole minutes and exact seconds. Watching the flight telemetry timer on an LCD transmitter display requires setting the alarm to an exact MM:SS boundary before touchdown.

Extract whole minutes and convert the fractional portion: 11 [ENTER] .52 [ENTER] 60 [×] $\to$ 11 min 31 sec (T: 0, Z: 0, Y: 11, X: 31.2). Multiplying the $0.52\text{ min}$ decimal fraction by 60 reveals that the safe mission timer should be set to $\mathbf{11\text{ minutes and } 31\text{ seconds}}$.

Workbook Challenge 9.17: Solar Tracker Bracket: Power at $30^\circ$ Morning Tilt ($10\text{ W peak} \times 0.866$)

You 3D-print an articulated mounting bracket to align a mini solar panel toward the morning sun. When the sun is oriented at a $30^\circ$ incident angle relative to the panel normal, Lambert's Cosine Law reduces the effective solar flux factor to $\cos 30^\circ \approx 0.866$. The panel carries a rated peak output of $10\text{ W}$ under perpendicular light.

Calculate actual morning electrical power with 10 [ENTER] .866 [×] $\to$ 8.66 Watts (T: 0, Z: 0, Y: 0, X: 8.66). Multiplying peak power by the projection coefficient shows the panel delivers $\mathbf{8.66\text{ Watts}}$ at a $30^\circ$ tilt angle.

Workbook Challenge 9.18: Solar Tracker Bracket: Power at $45^\circ$ Autumn Tilt ($10\text{ W peak} \times 0.707$)

During late autumn afternoon testing, the sun drops lower in the sky to an oblique $45^\circ$ angle of incidence ($\cos 45^\circ \approx 0.707$). Measuring the output current and voltage with an inline USB tester plugged into the charging port confirms the reduced solar harvesting rate.

Compute the lower sun angle generation with 10 [ENTER] .707 [×] $\to$ 7.07 Watts (T: 0, Z: 0, Y: 0, X: 7.07). The calculation reveals that output drops to $\mathbf{7.07\text{ Watts}}$ at $45^\circ$, demonstrating why automated two-axis solar tracking increases daily energy capture.

Workbook Challenge 9.19: Outfitting the Workshop: Tool Backpack $30\%$ Discount ($\$56.00$)

To carry your SC-6 calculator, calipers, multimeter, and 3D printing tools between home and school, you select an engineering field backpack with a regular retail price of $\$56.00$, currently on sale for $30\%$ off. Laying out play money or counting bills on the workbench helps visualize the $30\%$ savings before walking to the hardware register.

In RPN, tapping [%] computes the discount dollar savings while preserving the base price in Register $Y$, allowing a single subsequent tap of [-] to find the net discounted price: 56 [ENTER] 30 [%] [-] [$] $\to$ $39.20 (T: 0, Z: 0, Y: 0, X: 39.2). The stack reveals the discounted sale price of $\mathbf{\$39.20}$ before taxes.

Workbook Challenge 9.20: Workshop Equipment Sales Tax ($7.5\%$ on $\$39.20$)

At the cash register, state and municipal sales tax of $7.5\%$ is applied to your $\$39.20$ discounted backpack purchase. Inspecting a printed store receipt shows how sales tax adds a fractional charge to every dollar spent.

Evaluate the final receipt total with 39.20 [ENTER] 7.5 [%] [+] [$] $\to$ $42.14 (T: 0, Z: 0, Y: 0, X: 42.14). Tapping [%] computes the exact tax amount ($\$2.94$) while keeping $\$39.20$ in Register $Y$, and tapping [+] totals the purchase at $\mathbf{\$42.14}$.

Workbook Challenge 9.21: Team Material Budget Split ($\$124.80 \div 6$ members)

Six student builders in your school robotics club jointly order a shared assortment of stainless ball bearings, pulleys, and stepper motors totaling $\$124.80$. To settle the group hardware invoice fairly, you divide the total expense into six equal shares so every team member contributes their exact portion.

Calculate the individual contribution using 124.80 [ENTER] [6] [÷] [$] $\to$ $20.80 (T: 0, Z: 0, Y: 0, X: 20.8). Formatting the result in currency mode confirms that each student maker owes $\mathbf{\$20.80}$ toward the team parts kit.

Workbook Challenge 9.22: Interlocking Snap Cube Sliced Volume ($2\text{ cm} \times 2\text{ cm} \times 2\text{ cm}$)

You are modeling custom interlocking snap cubes for classroom math manipulatives. Holding a physical plastic snap cube in your hand and measuring each of its three square-cornered edges with a millimeter ruler confirms dimensions of exactly $2.0\text{ cm} \times 2.0\text{ cm} \times 2.0\text{ cm}$.

Compute the solid cube volume with [2] [ENTER] [2] [×] [2] [×] $\to$ 8.0 cm³ (T: 0, Z: 0, Y: 0, X: 8). Multiplying base area by height on the stack confirms an overall volume of exactly $\mathbf{8.0\text{ cm}^3}$.

Workbook Challenge 9.23: Snap Cube Theoretical Solid Mass ($\rho = 1.24\text{ g/cm}^3$)

If you were to 3D-print the $8.0\text{ cm}^3$ snap cube completely solid ($100\%$ infill) out of standard corn-based PLA thermoplastic ($\rho = 1.24\text{ g/cm}^3$), what would the block weigh? Placing an eight-cube block array on your balance pan provides a physical feel for solid plastic density before optimizing the internal slicer infill.

Calculate theoretical solid mass with [8] [ENTER] 1.24 [×] $\to$ 9.92 grams (T: 0, Z: 0, Y: 0, X: 9.92). Multiplying volume by density confirms that a $100\%$ solid snap cube weighs $\mathbf{9.92\text{ grams}}$.

Workbook Challenge 9.24: Material Savings with $20\%$ Gyroid Infill ($3.1\text{ g actual}$)

To conserve filament and speed up batch fabrication, you slice the snap cube with lightweight $20\%$ gyroid internal infill and 2 solid perimeter shells. Placing the printed cube on your sensitive digital scale reveals that it weighs only $3.1\text{ grams}$. Comparing the printed cube against the $9.92\text{ g}$ solid benchmark on a two-pan balance highlights the significant material savings.

Find the mass of filament saved with 9.92 [ENTER] 3.1 [-] $\to$ 6.82 g saved (T: 0, Z: 0, Y: 0, X: 6.82). The subtraction confirms that using $20\%$ gyroid infill saves $\mathbf{6.82\text{ grams}}$ of plastic per block while preserving structural rigidity.

Workbook Challenge 9.25: Solar Cooker Thermal Dissipation in Cloud Shade ($70^\circ \to 40^\circ\text{C}$ in $20\text{ min}$)

While testing thermal insulation materials on your solar cooker box, a dense afternoon storm cloud covers the sun. Logging the cooker's internal temperature drop over a $20\text{-minute}$ shading period shows the absorber cooling from $70^\circ\text{C}$ down to $40^\circ\text{C}$. Plotting the data points on graph paper generates a smooth exponential cooling curve.

Compute the rate of thermal dissipation with 70 [ENTER] 40 [-] 20 [÷] $\to$ 1.50 °C/min (T: 0, Z: 0, Y: 0, X: 1.5). Subtracting temperatures yields a $30^\circ\text{C}$ thermal loss, and dividing by 20 minutes indicates a steady cooling rate of $\mathbf{1.50^\circ\text{C/min}}$.

Workbook Challenge 9.26: Quarter-Batch Polymer Clay Recipe ($3\ \frac{1}{2}\text{ cups milk} \div 4$)

You are mixing a small experimental batch of homemade casein polymer clay using warm milk and vinegar to cast custom pulley wheels. The master studio recipe calls for $3\ \frac{1}{2}$ cups of milk, but you only want to mix a $1/4$ test batch. Filling an 8-fluid-ounce measuring cup to the 7-ounce mark provides a physical benchmark for $7/8\text{ cup}$.

Calculate the scaled liquid volume using mixed fraction division: 3 [.] 1 [.] 2 [ENTER] [4] [÷] [a ᵇ/c] $\to$ 7/8 cup (T: 0, Z: 0, Y: 0, X: 0.875). Tapping [a ᵇ/c] formats the result into a proper fraction, confirming that the quarter-batch requires exactly $\mathbf{7/8\text{ cup}}$ of milk.