Hey Goldfinger,

Really good question here - thanks for sharing! To answer this one, think about what we know has to be true of n:

n! (the product of all integers from 1 to n) MUST BE a multiple of 990.

What does it take to be a multiple of 990? When you're dealing with factors and multiples, you'll typically want to break down numbers into prime factors - prime factors tell you the essential components of a number like 990, almost like breaking a substance down into its atomic components or something like that (water = H20 which means you need 2 hydrogen atoms for every one oxygen atom - the process of prime factorization is similar...it breaks numbers down to their fundamental components).

990 factors out to 99*10, which gets us the prime factors:

99 = 3*3*11
10 = 2*5

So for our number to be a multiple of 990, it MUST have the following prime factors: 2*3*3*5*11

In order to get that 11, the absolute minimum value of n has to be 11, because as a prime number 11 isn't going to be constructed by multiplying any two other integers together. So if n is 11, then n! will contain:

1*2*3*4*5*6*7*8*(3*3)*10*11

(note that we can break that 9 down into 3*3 to ensure that we have that second 3 covered)

n = 11 satisfies all of our requirements to be a multiple of 990, and we know that it's the lowest such number because of that necessary 11. Therefore, the answer is 11.

Thanks Brian,

looks easy, is easy, but somehow in the test I did get it wrong. Hope this doesn't happen in the real GMAT....