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# C&P Problem

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Two distinct positive integers from 1 to 50 inclusive are chosen. Let the sum of the integers equal S and the product equal P. What is the probability that P + S is one less than a multiple of 5?

May 14, 2024

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There are 50⋅49=2450 total possible choices of two distinct positive integers from 1 to 50 inclusive. We want the sum P+S to leave a remainder of 4 when divided by 5.

We can achieve this by looking at even and odd combinations. If P is even, then S must be odd for P+S to be odd. If P is odd, then S must be even for P+S to be odd.

Enumerate through all the even numbers from 2 to 50 inclusive: For each even number i, enumerate through all the odd numbers from 1 to 49 inclusive and check if i⋅j+i+j≡4(mod5) (i.e. leaves a remainder of 4 when divided by 5). If it does, then this is a favorable outcome.

There are 25 even numbers from 2 to 50 inclusive, and 25 odd numbers from 1 to 49 inclusive so this gives us 25⋅25=625 total possibilities to check (though many will be redundant).

We compute that there are 225 favorable outcomes.

The probability is then $\frac{225}{2450} = \boxed{\frac{9}{98}}$.

May 26, 2024