jn0r wrote:
What is a remainder when 15! is divided by 1001?
A) 0
The answer really has to be zero here, because if it was anything else, there's no GMAT-level math you could use to find the answer. For example, if a question asked for the remainder when 12! is divided by 13, that might look like a simpler question because the numbers are smaller, but it's not, because 12! is not divisible by 13. You'd need to use modular arithmetic to answer this question, and you don't need math that advanced on the test.
As bidskamikaze correctly points out, 1001 = (7)(11)(13), and 15! is divisible by (7)(11)(13). I'd add though that:
bidskamikaze wrote:
Important stuff to remember: 1001 = 7 x 11 x 13this "1001" appears frequently on GMAT, so better to remember it
I would be very surprised to see a real GMAT question which required you to prime-factorize 1001 from scratch. If a GMAT question is going to ask you to prime factorize a large number (bigger than 200), the number really needs to have an obvious divisor (2, 3 or 5, or a divisor obvious for some other reason), and 1001 does not have any obvious divisors if you haven't seen the number before. In theory, prime factorizing numbers that large can take a very long time if there are no obvious divisors, which is why it's not something the GMAT can test.