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2021 WSMO Accuracy Round Problems/Problem 1

Problem

Find the sum of all the positive integers $n$ such that $n$ is $\frac{2n^2-5n+5}{n-5}$ an integer.

Solution

Note that $2n^2-5n+5=(n-5)(2n+5)+30.$ The given condition implies that $\frac{(n-5)(2n+5)+30}{n-5}=2n+5+\frac{30}{n-5},$ which implies that $n-5$ is a factor of 30. Thus, \[n-5=-30,-15,-10,-6,-5,-3,-2,-1,1,2,3,5,6,10,15,30.\] Solving, we find that \[n=-25,-10,-5,0,2,3,4,6,7,8,10,11,15,20,35.\] Since $n$ is positive, we find that $n=2,3,4,6,7,8,10,11,15,20,35.$ Summing, we find that the answer is \[2+3+4+6+7+8+10+11+15+20+35=\boxed{121}.\]

~pinkpig

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