Difference between revisions of "2021 Fall AMC 12B Problems/Problem 16"
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
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Revision as of 02:10, 27 November 2021
Contents
[hide]Problem
Suppose , , are positive integers such that and What is the sum of all possible distinct values of ?
Solution 1
Let , , . WLOG, let . We can split this off into cases:
: let we can try all possibilities of and to find that is the only solution.
: No solutions. By and , we know that , , and have to all be divisible by . Therefore, cannot be equal to .
: C has to be both a multiple of and . Therefore, has to be a multiple of . The only solution for this is .
: No solutions. By and , we know that , , and have to all be divisible by . Therefore, cannot be equal to .
: No solutions. By and , we know that , , and have to all be divisible by . Therefore, cannot be equal to .
: No solutions. By and , we know that , , and have to all be divisible by . Therefore, cannot be equal to .
: No solutions. As , , and have to all be divisible by , has to be divisible by . This contradicts the sum .
Putting these solutions together, we have
-ConcaveTriangle
Solution 2
Because is odd, , , are either one odd and two evens or three odds.
: , , have one odd and two evens.
Without loss of generality, we assume is odd and and are even.
Hence, and are odd, and is even. Hence, is even. This violates the condition given in the problem.
Therefore, there is no solution in this case.
: , , are all odd.
In this case, , , are all odd.
Without loss of generality, we assume
: , , .
The only solution for is (7, 9, 7).
Hence, .
: , , .
The only solution for is (5, 3, 15).
Hence, .
: , , .
There is no solution in this case.
Therefore, putting all cases together, the answer is .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 12 Problems and Solutions |
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