OG13 - PS - Q110 - is that really obvious?

I definitely don't think it's obvious to non-GMAT-experts! The trick is getting yourself to think like an expert, though...

Any positive integer can be expressed as the product of prime factors - we call this PRIME FACTORIZATION. For example, the prime factorization of 140 is (2^2)(5^1)(7^1), and 108 = (2^2)(3^3). This prime factorization is really helpful when you're solving questions with variable exponents. For example:

12 = (2^x)(3^y)
In order to solve for x and y here, we need to break 12 down to its prime factors:
12 = (2^2)(3)
So, (2^2)(3) = (2^x)(3^y)
Therefore x = 2 and y = 1.

In #110, we have "k" and "k + 1" as exponents, which is a big clue that we probably want to think in terms of prime factorization.

The hardest part about this question is simply figuring out what it's asking - breaking down "a^k is a divisor of b, but a^(k + 1) is not." How do we translate that? So "a" to a certain exponent goes evenly into "b," but "a" to the next highest exponent does not. That would mean that a^k was the maximum number of a's that go into "b." In other words... all the a's!

So, 2^k || 72 would be the maximum number of factors of 2 that go into 72 - all the 2's in 72. If we factor 72, we find that the prime factorization is (2^3)(3^2). There are 3 factors of 2 in 72 (that's what 2^3 tells us), so k must equal 3.

There's more on prime factorization here: https://www.beatthegmat.com/2-x-2-x-2-3- ... tml#578272