Difference between revisions of "2000 AMC 10 Problems/Problem 14"
(New page: 71, 76, 80, 82, 91. The sum of the first 2 must be even, so we must choose 2 evens or the 2 odds. Let us look at the numbers (mod 3). 2,1,2,1,1. If we choose the two odds, the next num...) |
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Revision as of 19:52, 8 January 2009
Problem
Solution
71, 76, 80, 82, 91.
The sum of the first 2 must be even, so we must choose 2 evens or the 2 odds.
Let us look at the numbers (mod 3).
2,1,2,1,1.
If we choose the two odds, the next number must be a multiple of 3, of which there is none.
Similarly, if we choose 76,80 or 80,82, the next number must be a multiple of 3, of which there is none.
So we choose 76,82 first.
The next number must be 1 (mod 3), of which only 91 remains.
The sum of these is 249. This is equal to 1 (mod 4).
Thus, we need to choose one number that is 3 (mod 4). 71 is the only one that works.
Thus, 80 is the last score entered.
C.
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
