Difference between revisions of "2004 AMC 10A Problems/Problem 7"
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Therefore, there are <math>5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100</math> oranges in the stack <math>\Rightarrow\mathrm{(C)}</math>. | Therefore, there are <math>5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100</math> oranges in the stack <math>\Rightarrow\mathrm{(C)}</math>. | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 01:39, 11 September 2007
Problem
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?
Solution
There are 
oranges on the 1st layer of the stack. When the 2nd layer is added on top of the first, it will be a layer of 
oranges. When the third layer is added on top of the 2nd, it will be a layer of 
oranges, etc.
Therefore, there are 
oranges in the stack 
.
See also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
