AMC 10 · 2004 · #7
Grade 4 geometry-3dA grocer stacks oranges in a pyramid-like stack whose rectangular base is 5 oranges by 8 oranges. Each orange above the first level rests in a pocket formed by four oranges below. The stack is completed by a single row of oranges. How many oranges are in the stack?
Pick an answer.
AMC 10 2004 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: Oranges are stacked so the bottom layer is a $5$-by-$8$ rectangle. Each orange in a higher layer sits in the pocket made by four oranges below it. The stack ends with a single straight row of oranges. Find the total number of oranges in the whole stack.
Givens: The bottom layer is a rectangle of oranges, $5$ oranges by $8$ oranges.; Each orange above the bottom rests in a pocket formed by four oranges just below it.; The stack is finished off by one single row of oranges at the top.; Answer choices: (A) $96$, (B) $98$, (C) $100$, (D) $101$, (E) $134$
Unknowns: The total number of oranges across all layers of the stack.
Understand
Restated: Oranges are stacked so the bottom layer is a $5$-by-$8$ rectangle. Each orange in a higher layer sits in the pocket made by four oranges below it. The stack ends with a single straight row of oranges. Find the total number of oranges in the whole stack.
Givens: The bottom layer is a rectangle of oranges, $5$ oranges by $8$ oranges.; Each orange above the bottom rests in a pocket formed by four oranges just below it.; The stack is finished off by one single row of oranges at the top.; Answer choices: (A) $96$, (B) $98$, (C) $100$, (D) $101$, (E) $134$
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #5 Look for a Pattern, #7 Identify Subproblems
The whole problem hinges on seeing the geometry (Tool #17): an orange resting in a pocket of four sits over the gap where four oranges meet, so it lands between rows and between columns of the layer below. That means each new layer has one fewer orange along each side than the layer under it. Once you see that, the sizes form a clear shrinking pattern (Tool #5): $5\times8$, then $4\times7$, then $3\times6$, and so on. Finally the stack splits into simple subproblems (Tool #7): count each rectangular layer by multiplying its two side lengths, then add the layers. The trap answer is to forget how the stack ends or to miscount the last layer.
Execute — Answer: C
4.OA.C.5 Step 1 See why each layer is one smaller on each side
- Picture the oranges below as a grid of rows and columns.
- An orange in the next layer up drops into a pocket where four oranges meet, so it sits over a gap, centered between two rows and between two columns.
- The number of gaps along a side of length $m$ is $m-1$, and along a side of length $n$ is $n-1$.
- So a layer that is $m$ by $n$ supports a layer that is $(m-1)$ by $(n-1)$: one fewer along each direction.
💡 An orange balances on the gap between four others, and there is always one fewer gap than there are oranges in a row.
4.OA.C.5 Step 2 List the size of every layer
- Start from the $5$-by-$8$ base and drop each side by $1$ for the next layer.
- The sizes are $5\times8$, $4\times7$, $3\times6$, $2\times5$, and $1\times4$.
- The shorter side counts down $5,4,3,2,1$, so it hits $1$ after five layers, and $1\times4$ is exactly the single finishing row the problem describes.
- That confirms the stack has five layers and no more.
💡 Follow the shrink rule until a side reaches $1$, which is the single row that finishes the stack.
3.OA.A.3 Step 3 Count the oranges in each layer
- Each layer is a rectangle, so its orange count is the product of its two side lengths.
- Multiply each pair: $5\times8=40$, $4\times7=28$, $3\times6=18$, $2\times5=10$, and $1\times4=4$.
- These five numbers are the oranges on each level from bottom to top.
💡 A rectangular grid of oranges is just rows times columns.
3.NBT.A.2 Step 4 Add the layers for the total
- Add the five layer counts: $40+28+18+10+4$.
- Grouping to make it easy, $40+28=68$, then $68+18=86$, then $86+10=96$, then $96+4=100$.
- So the stack holds $100$ oranges, which is choice (C).
💡 The total is just the sum of every layer's count.
4.OA.C.5 Picture the oranges below as a grid of rows and columns. An orange in the next l 4.OA.C.5 Start from the $5$-by-$8$ base and drop each side by $1$ for the next layer. The 3.OA.A.3 Each layer is a rectangle, so its orange count is the product of its two side le 3.NBT.A.2 Add the five layer counts: $40+28+18+10+4$. Grouping to make it easy, $40+28=68$ Review
Reasonableness: The base alone is $5\times8=40$ oranges, and each higher layer is smaller, so the total must be more than $40$ but not by a huge amount. Five shrinking layers landing at $100$ fits: it is more than double the base but far below $134$, which would require far more or larger layers. It is also just above $96$ (the base plus a rough guess), and the careful layer-by-layer sum lands cleanly on $100$, so (C) is sensible while (E) is clearly too big.
Alternative: Instead of listing sizes, use the shrink rule as a formula: the layers are $(6-k)\times(9-k)$ for $k=1,2,3,4,5$. Evaluating gives $5\cdot8,4\cdot7,3\cdot6,2\cdot5,1\cdot4 = 40,28,18,10,4$, and summing again yields $100$. Writing it as a single expression $\sum_{k=1}^{5}(6-k)(9-k)$ confirms the same total without drawing the stack.
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Turning the pocket-of-four geometry into the rule that each layer is one smaller on each side, then generating the sizes $5\times8, 4\times7, 3\times6, 2\times5, 1\times4$.)3.OA.A.3Solve multiplication and division word problems within 100 (Counting each rectangular layer as its number of rows times its number of columns.)3.NBT.A.2Fluently add and subtract within 1000 (Adding the five layer counts $40+28+18+10+4$ to get the total of $100$.)
⭐ Each orange sits in the gap between four below it, so every layer is one smaller on each side; multiply each layer's sides and add: 40+28+18+10+4 = 100.
⭐ Each orange sits in the gap between four below it, so every layer is one smaller on each side; multiply each layer's sides and add: 40+28+18+10+4 = 100.
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