2014 AMC 12B Problems/Problem 25
Problem
Find the sum of all the positive solutions of
Solution 1
Rewrite as . Now let , and let . We have:
Therefore, .
Notice that either and or and . For the first case, only when and is an integer. when is an even multiple of , and since , only when is an odd divisor of . This gives us these possible values for : For the case where , , so , where m is odd. must also be an odd multiple of in order for to equal , so must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for , and therefore no cases where and . Therefore, the sum of all our possible values for is
Solution 2
Very similar to the solution above, re-write the expression using :
Now, expand the LHS and cancel terms:
Now we use product-to-sum identities to get:
Notice that for any , . This is achieved when , or equivalently
We can cleverly assume for some real . Then, we must have
In order for this to be satisfied, must be an even integer. Factoring , we see that must be a positive odd integer that divides . Our only positive valid are . Our answer is just .
-FIREDRAGONMATH16
Solution 3
Rewriting as and transposing from the LHS to the RHS, we get,
By the AM-GM Inequality,
Also, because of the range of ,
Hence, , and we get (),
From and ,
Therefore, sum of values of is .
~ plusone
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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