2004 AMC 10A Problem 7

1. Picture one orange resting on four oranges below it. It sits over the gap between two neighboring rows and two neighboring columns.
2. Between $5$ rows there are $4$ gaps, and between $8$ columns there are $7$ gaps. So the layer above a $5 \times 8$ layer is $4 \times 7$. Each new layer loses one in each direction.
3. Keep going until only one row is left:

The top layer is a single row of $4$, just as the problem says.
4. Add the layers:
$$40 + 28 + 18 + 10 + 4 = 100.$$

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

The phrase "rests in a pocket formed by four oranges" is a rule for building the next layer: both sides drop by one. With only five layers, writing them all out is faster than any formula, and the table also shows clearly where the stack has to stop.

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 $\boxed{\textbf{(C)}\ 100}$.