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This math problem comes from a recent Mathcounts competition.

"what is the sum of the positive integers p for which the value of 13/(p^2-3) is a positive integer?"

An eighth grade student answered this question within 30 seconds. It took me much longer.

To solve this problem, you have to realize that they are asking for cases in which p is an integer and 13/(p^2-3) is a positive integer.

Setting the denominator to zero, we find two discontinuities: 3^(1/2) and -3^(1/2). Taking the positive, let's find out if the curve 13/(p^2-3) is positive or negative as we approach 3^(1/2). BTW: 3^(1/2) is approximately 1.73.

Let's try p=1. We get 13/(p^2-3) equaling -13/3. In fact, closer we get to 1.73, the more negative this curve gets. Therefore this problem is not interested in an value of p smaller than 1.73. This problem is also not interested in any p in which 13/(p^2-3) is less than one. Solving this, we get 4.

Therefore, the values of p of interest (again, only integers) are 2, 3, and 4. Plugging in 2, we get 13, this is an integer. Plugging in 3 we get 13/6, which is not an integer. Plugging in 4, like we did above, we get 1.

To solve this problem, we then go: 4+2=6.

The answer is six. Not quite 30 seconds, but not too shabby.