+86 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?
+180 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}}$.
