Hi melanie.espeland,

Yes, you are correct! For a number to be a multiple of 990, that number must have the exact same prime factors (including duplicates) as 990 (but may have "extra" prime factors as well).

This prompt adds the extra stipulation that N has to be as SMALL as possible, so after factoring 990, we need to find the smallest product that "holds" the 2, 5, 11, and two 3s that make up 990.

1(2)(3)(4)....(11) is the smallest product that does that, so N = 11.

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990. What is least possible value of n?
a) 10
b) 11
c) 12
d) 13
e) 14

A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N
Similarly, we can say:
If N is is a multiple of k, then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7

So, if if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.

Since 990 = (2)(3)(3)(5)(11), we know that one 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.

For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.
So, the answer is B

Cheers,
Brent

Hi j_shreyans,

This series of posts provides a nice explanation of the "math" involved and the steps necessary to answer the question. Was there anything about it that you specifically want to know more about?

GMAT assassins aren't born, they're made,
Rich