You can see why this works if you build factor tables for a regular number and a perfect square:
Here all the factors come in pairs, so we will have an even number of factors.
Here there is a pair of identical factors (that's the perfect square part), so we will always have an odd number.
Therefore, statement 1 tells us we don't have a perfect square. Our answer is "no"--sufficient.
As for the sum of the distinct factors, it's a bit more theoretical, but let's look at two possibilities:
1) The number is odd, in which case all of its factors are odd. If they come in pairs, each pair will add to make an even. (O+O=E) If the number is a perfect square, it will have an odd number of factors, leaving an extra odd number, which will make an odd sum. (E+O=O) Therefore, the sum of an odd perfect square's factors must be odd.
2) The number is even. If it is also a perfect square, it will have an even number of each prime factor, because the primes have to pair off (e.g. 100 = 10*10 = (2*5)(2*5)). All of its factors aside from 1 will be built from these primes, we will have an even number of evens (adding up to an even) and an even number of odds (also adding up to an even). Then 1 comes in and messes up all that harmony and makes the whole thing odd! Therefore, the sum of an even perfect square's factors must be odd.
So in either case, a perfect square's factors make an odd sum. Statement 2 gives an answer of "no," which is sufficient. The answer is D.
I hope this helps! You can definitely try this with some different numbers--it makes more sense when you play with a few examples.
Dmitry Farber | Manhattan GMAT Instructor | New York
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