Is the positive integer

*N*a perfect square?

(1) The number of distinct factors of

*N*is even.

(2) The sum of all distinct factors of

*N*is even.

Answer Explanation:

(1) SUFFICIENT: The factors of any number

*N*can be sorted into pairs that multiply to give

*N*. (For instance, the factors of 24 can be paired as follows: 1 and 24; 2 and 12; 3 and 8; 4 and 6.) However, if

*N*is a perfect square, one of these ‘pairs’ will consist of just one number: the square root of

*N*. (For example, if

*N*were 49, it would ahve the factor pair 7 × 7) Since all of the other factors can be paired off, it follows that if

*N*is a perfect square, then N has an odd number of factors. (If

*N*is not a perfect square, then all of its factors can be paired off, so it will have an even number of factors.) This statement then implies that

*N*is not a perfect square.

(2) SUFFICIENT: Let

*N*be a perfect square. If

*N*is odd, then all factors of

*N*are odd. Therefore, by the above reasoning,

*N*has an odd number of odd factors, so their sum must be odd. If

*N*is even, let

*M*be the product of all the odd prime factors (as many times as they appear in

*N*– not distinct) of

*N*, which is also a perfect square. (For instance, if

*N*= 100, then

*M*=5 × 5 = 25.) Then the sum of factors of

*M*is odd, by the above reasoning. Furthermore, all other factors of

*N*(i.e., that don’t also divide

*M*) are even. The sum total is thus odd + even = odd.

Therefore, the sum of the factors of any perfect square is odd, so this statement implies that

*N*is not a perfect square.

This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

The correct answer is D.

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I picked numbers to get to my answer, but made a silly mistake by omitting "1" as a distinct factor in Statement II.

And to the MGMAT explanation - hunh?!

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