If x represents the number of positive factors of integer y

[BTGmoderatorDC]

If x represents the number of positive factors of integer y, is x odd?

(1) y = n! where n is a positive integer greater than 1
(2) y = m^2 − 1 where m is a positive integer greater than 1

OA D

Source: Veritas Prep

[Ian Stewart]

A number has an odd number of divisors only if that number is a perfect square. It's easy to see why: for a number that is not a perfect square, say 6, then all of the divisors can be 'paired up' into pairs that produce 6 as a product:

1 and 6
2 and 3

so we have an even number of divisors. But for a perfect square, there will always be one divisor, the square root of the number, that is not in a pair, so for, say, 16 we have

1 and 16
2 and 8
4

and thus have an odd number of divisors.

So the question is just asking "is y a perfect square?" Statement 2 then immediately proves y is not a perfect square, since y is one less than some perfect square, and there are no two perfect squares that are one apart (besides 0 and 1). So Statement 2 is sufficient to give a 'no' answer to the question.

Statement 1 is also sufficient, since n! is never a perfect square if n > 1. But if you can prove that using only GMAT level math (i.e. without using something known as "Bertrand's postulate", which you definitely do not need to know for the GMAT) then you probably have a paper you could publish in an advanced math journal. So while Statement 1 is technically sufficient, there's no way a GMAT-level test taker would be able to demonstrate that, and the question is very far out of the scope of the test.