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2004 AMC 10A Problem 7

2004 AMC 10A Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 2 of 5 · about 2 min · position P1-10

A 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?

A)
$96$
B)
$98$
C)
$100$
D)
$101$
E)
$134$

Answer

Key insight

Each layer loses one orange in each direction, so the layers are 5x8, 4x7, 3x6, 2x5, 1x4 and the stack stops at the single row of 4.

Solution

An orange sitting in a pocket formed by four oranges lies over the gap between two rows and two columns. So a layer above an $m \times n$ rectangle is an $(m-1) \times (n-1)$ rectangle.

Starting from $5 \times 8$ , the layers are
$5 \times 8,\quad 4 \times 7,\quad 3 \times 6,\quad 2 \times 5,\quad 1 \times 4.$
The last layer is a single row of $4$ , matching the statement that the stack ends in one row.

Adding the layer sizes:
$40 + 28 + 18 + 10 + 4 = 100.$

The answer is $\boxed{\textbf{(C)}\ 100}$ .

Why this works

A physical description ("rests in a pocket formed by four oranges") encodes a recursion: both dimensions drop by one per level. Write out the terms explicitly rather than reaching for a formula; the sum has only five terms, and listing them also shows exactly where the stack must stop.

The trap

Stopping the stack at 2x5 or continuing past the 1x4 row, giving 96 or an impossible extra layer.

Common mistakes

Stopping the stack at 2x5 or continuing past the 1x4 row, giving 96 or an impossible extra layer.
Assuming each layer shrinks in only one dimension (for instance $5\times8, 5\times7, \dots$ ).

Techniques

Set up the equation/formula and compute; no special trick needed · Compute small cases, spot the pattern, generalize