2004 AMC 10A Problem 7
A grocer stacks oranges in a pyramid-like stack whose rectangular base is oranges by 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?
Show solution without answering
Key insight
Each orange sits in a gap between four below, so each layer shrinks by one both ways: 5 by 8, 4 by 7, down to 1 by 4.
Solution
- Picture one orange resting on four oranges below it. It sits over the gap between two neighboring rows and two neighboring columns.
- Between rows there are gaps, and between columns there are gaps. So the layer above a layer is . Each new layer loses one in each direction.
- Keep going until only one row is left:
| layer | size | oranges |
|---|---|---|
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | ||
| 5 |
The top layer is a single row of , just as the problem says.
4. Add the layers:
The answer is .
Why this works
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.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
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 rectangle is an rectangle.
Starting from , the layers are
The last layer is a single row of , matching the statement that the stack ends in one row.
Adding the layer sizes:
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
- Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
- Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)
- Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic (§12.6)
Similar problems
Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.