2021 Fall AMC 12B Problems/Problem 16
Contents
Problem
Suppose , , are positive integers such that and What is the sum of all possible distinct values of ?
Solution 1 (Observation)
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 is .
Hence, .
: , , .
The only solution is .
Hence, .
: , , .
There is no solution in this case.
Therefore, putting all cases together, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 2 (Enumeration)
Let , , . Without the loss of generality, 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 .
: Note that 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 3
Since , or .
As , it is impossible for , so .
This means that , , and must all be coprime. The only possible ways for this to be true are and .
Without loss of generality, let . Since , then or .
.
~bkunzang
Video Solution By Power Of Logic
~~Hayabusa1
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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All AMC 12 Problems and Solutions |
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