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There is a formula that lets you find the sum of all of a number's divisors from a prime factorization, which I'm sure you'll find by googling a suitable search phrase. You will *never* need that formula on the GMAT, so I won't post it here. If you don't have a formula, you can still find the sum of a number's divisors if you know how to list the divisors systematically. Before I work through an example, I'd stress that this is for interest only; you won't see a question about this on the GMAT unless it's straightforward to just list the divisors and add, as in the example in the original post above.

Let's take a more difficult number, say 288 = (2^5)(3^2). What are the divisors of this number? Well we can start by ignoring the 3's; we have:

1, 2, 2^2, 2^3, 2^4, 2^5

Notice the sum of these, which I'll call S, is equal to 63.

We will then get another set of six divisors by multiplying each number in the list above by 3^1, and another set by multiplying each number in the list above by 3^2:

3, 3*2, 3*2^2, 3*2^3, 3*2^4, 3*2^5

Notice the sum of this list is just 3*S (that is, it's 3 times the sum of our first list).

3^2, 3^2 * 2, 3^2 *2^2, 3^2 * 2^3, 3^2 * 2^4, 3^2 * 2^5

Notice the sum of this list is just (3^2)*S = 9*S.

We've just listed all 18 of the divisors of 288. Their sum is just S + 3S + 9S = 13*S = 13*63.

The formula you can use here just generalizes the approach above (the above becomes a bit awkward if you have a lot of different prime divisors).

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