2004 AMC 10A Problem 7

Attempt Problem 7 of the 2004 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

A grocer stacks oranges in a pyramid-like stack whose rectangular base is 55 oranges by 88 oranges. Each orange above the first level rests in a pocket formed by four oranges in the level below. The stack is completed by a single row of oranges. How many oranges are in the stack?
9696
9898
100100
101101
134134
Answer: C
Concepts:basic countingsummation
Small hint:
Each layer up has one fewer orange in each dimension than the layer below
Big hint:
Sum 5⋅8+4⋅7+3⋅6+2⋅5+1⋅45\cdot 8 + 4\cdot 7 + 3\cdot 6 + 2\cdot 5 + 1\cdot 4
Solution:
There are five layers, each one shorter and narrower than the one below. The total number of oranges is 5⋅8+4⋅7+3⋅6+2⋅5+1⋅4=40+28+18+10+4=100. \begin{aligned} &5\cdot 8 + 4\cdot 7 + 3\cdot 6 \\ &\quad {}+ 2\cdot 5 + 1\cdot 4 \\ &= 40 + 28 + 18 + 10 + 4 = 100. \end{aligned}
Thus, the correct answer is C.
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