AegisAshore wrote:
An additional follow up to this question. I understand the solution presented and the fact that the sum of all distinct factors of N are odd when N is a perfect square. However, what confused me is that I tried an opposite scenario:
When N is NOT a perfect square. In this case it seemed as if the sum of all distinct factors could be EITHER even or odd. Hence, before seeing the solution, I thought (2) was not sufficient.
i.e.
Not a perfect square = 8: 1+2+4+8=15 = odd
Not a perfect square = 10: 1+2+5+10 = 18 = even
Does this need to be considered?
You're certainly correct that if you sum the divisors of non-squares, you can get an even sum or an odd sum. Here though, we don't need to consider that fact. If we know the sum of our divisors is even, we know that our number
cannot be a perfect square, since the sum of the divisors of a square is always odd. Since we can be certain the answer to the question is 'no' from Statement 2 alone, it is sufficient.
I'd add that I see this question posted all the time, and I really don't like it. While one can prove that the sum of the divisors of a square is odd (I'll do it quickly below), no GMAT test taker could be expected to know that fact, and further, no GMAT test taker could reasonably be expected to
prove that fact within two minutes. That leaves only one viable approach to the question - picking a few numbers hoping to find a pattern. When statements are sufficient in real GMAT questions, there is always some
logical way to see that in a very short space of time - much less than two minutes, since there's always the other statement to analyze as well! You are never forced to 'guess' based on some pattern you find among small numbers.
Now, for interest only, since it won't be tested on the GMAT, we can prove that the sum of the divisors of a perfect square is odd. If we have a perfect square, we know that in its prime factorization, all the powers are even. Let's take an example:
n = (2^6)(3^4)(7^2)
is a perfect square (it is the square of (2^3)(3^2)(7^1)). Now to count this number's divisors, we add one to each power and multiply - our number n has 7*5*3 = 105 divisors. Note that when we add one to each power, we can only get odd numbers, since each power is even. So we find the total number of divisors by multiplying odd numbers only, and the number of divisors must therefore be odd.
Now, to count the number of odd divisors only, we can do the same thing, except we completely ignore the (2^k) part of the prime factorization. For our number n above, we can just pretend that the 2^6 is not there, and add one to the other powers and multiply. Our number n therefore has 5*3 = 15 odd divisors (and the other 90 divisors are all even). Notice that we have an odd number of odd divisors, for the same reason that we have an odd number of divisors in total; we are multiplying only odd numbers here.
Finally, when we sum the divisors of our perfect square, the even divisors will add up to some even number, say E. When we then add the odd divisors, we are adding an odd number of odd numbers to E, so the sum must be odd.
All of that said, GMAT test takers won't need to worry about doing this on test day, nor will it be helpful to memorize facts about summing divisors.
