AMC 10 · 2017 · #6
Easy mode Grade 5You have small blocks that are 2 in×2 in×1 in. You want to pack them into a box that is 3 in×2 in×3 in. The blocks cannot overlap and cannot stick out. What is the most blocks you can fit?
Pick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Solid blocks measure $2\text{ in}\times2\text{ in}\times1\text{ in}$. They are packed, without overlapping and without sticking out, into a $3\text{ in}\times2\text{ in}\times3\text{ in}$ box. The question asks for the greatest number of blocks that can fit.
Givens: Each block is a solid $2\times2\times1$ rectangular prism; The box is a $3\times2\times3$ rectangular prism; Blocks may be turned to any orientation but cannot overlap or poke outside; Answer choices: $3,\ 4,\ 5,\ 6,\ 7$
Unknowns: The largest number of blocks that fit inside the box
Understand
Restated: Solid blocks measure $2\text{ in}\times2\text{ in}\times1\text{ in}$. They are packed, without overlapping and without sticking out, into a $3\text{ in}\times2\text{ in}\times3\text{ in}$ box. The question asks for the greatest number of blocks that can fit.
Givens: Each block is a solid $2\times2\times1$ rectangular prism; The box is a $3\times2\times3$ rectangular prism; Blocks may be turned to any orientation but cannot overlap or poke outside; Answer choices: $3,\ 4,\ 5,\ 6,\ 7$
Plan
Primary tool: #14 Extreme Principle
Secondary: #8 Analyze the Units, #17 Visualize Spatial Relationships, #1 Draw a Diagram
The word "largest" makes this a maximum question, so Tool #14 (Extreme Principle) sets the strategy: find a ceiling no packing can beat, then show that ceiling is actually reachable. Tool #8 (Analyze the Units) supplies the ceiling — total block volume can never exceed the box's volume, and dividing the two volumes caps the count. A volume cap alone is not proof you can reach it, because solid blocks might not tile perfectly, so Tool #17 (Visualize Spatial Relationships) and Tool #1 (Draw a Diagram) build an actual arrangement that hits the cap. Ceiling plus matching construction pins the answer exactly.
Execute — Answer: B
5.MD.C.5 Step 1 Compute both volumes
- Volume of a box is length times width times height.
- The big box holds $3\times2\times3=18$ cubic inches.
- Each block holds $2\times2\times1=4$ cubic inches.
- These numbers say how much space there is to fill and how much each block uses.
💡 Multiplying the three side lengths counts how many unit cubes a prism holds.
5.NF.B.3 Step 2 Cap the count with volume
- The blocks sit inside the box without overlapping, so their volumes add up to at most the box's volume.
- That means the number of blocks is at most $18\div4=4.5$.
- You cannot fit half a block, so the count is a whole number no bigger than $4.5$, which is $4$.
💡 Total stuff inside can never exceed the container, so the box's volume sets a hard ceiling.
5.MD.C.4 Step 3 Build a packing that reaches 4
- A volume cap only proves $4$ is possible if a real arrangement exists, so build one.
- View the box as $3$ wide, $2$ tall, $3$ deep.
- Stand three blocks so each is $1$ wide, $2$ tall, $2$ deep; lined up across the width they fill the $3\times2\times2$ chunk in the front.
- That leaves a $3\times2\times1$ slab at the back.
- Lay a fourth block flat there as $2$ wide, $2$ tall, $1$ deep — it fits since the slab is $3$ wide and $2$ tall.
- That is $4$ blocks with no overlap and nothing sticking out.
💡 Filling the box in flat layers turns the 3-D puzzle into stacking simple rectangles.
5.MD.C.4 Step 4 State the maximum
- Volume forbids more than $4$ blocks, and the arrangement above fits exactly $4$.
- The two facts meet, so the largest number of blocks is $4$, which is choice $\textbf{(B)}$.
💡 When the ceiling and a real packing agree, that shared number is the exact answer.
5.MD.C.5 Volume of a box is length times width times height. The big box holds $3\times2\ 5.NF.B.3 The blocks sit inside the box without overlapping, so their volumes add up to at 5.MD.C.4 A volume cap only proves $4$ is possible if a real arrangement exists, so build 5.MD.C.4 Volume forbids more than $4$ blocks, and the arrangement above fits exactly $4$. Review
Reasonableness: Check the bookkeeping: $4$ blocks use $4\times4=16$ cubic inches inside an $18$-cubic-inch box, leaving $2$ cubic inches empty — a small $1\times2\times1$ gap in one corner, which is fine since no block can squeeze into a $1$-inch-thick leftover. A fifth block would need $20$ cubic inches of contents in an $18$-cubic-inch box, which is impossible. So $\textbf{(B)}\ 4$ is both achievable and the most that fits.
Alternative: Tool #3 (Eliminate Possibilities) trims the choices fast: any count of $5$ or more demands at least $20$ cubic inches but the box has only $18$, so $5,6,7$ are out immediately. That leaves $3$ or $4$, and since a concrete arrangement of $4$ exists, the larger survivor $4$ wins.
CCSS standards used (min grade 5)
5.MD.C.5Relate volume to the operations of multiplication and addition (Finding the box volume $3\times2\times3=18$ and the block volume $2\times2\times1=4$ by multiplying side lengths.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Dividing $18$ by $4$ to get $4.5$ and reading it as a whole-number cap of $4$ on the block count.)5.MD.C.4Measure volumes by counting unit cubes (Arranging blocks in flat layers to show that exactly $4$ blocks fill the box without overlap.)
⭐ Box volume $18$ divided by block volume $4$ is $4.5$, so at most $4$ blocks fit — and stacking them in flat layers really does fit $4$, choice $\textbf{(B)}$.
⭐ Box volume $18$ divided by block volume $4$ is $4.5$, so at most $4$ blocks fit — and stacking them in flat layers really does fit $4$, choice $\textbf{(B)}$.
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