yousufa wrote:
If n is a positive integer and the product of all the intergers from 1 to n, inclusive, is a multiple of 990, what is the 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 yousufa,

Finding the prime factors of a number is a pretty straight-forward idea: Find ANY number that goes into the larger number and then divide it out, break it down into primes, then continue repeating that process until you have a list of primes.

In the case of 990, you might spot that 2, 3, 5, 10, 99, etc. are divisible to start.

Let's say you divide out a 10:

Now you have 990 = 10 x 99

Break the 10 down into 2 x 5

Break the 99 down into 9 x 11

Break the 9 into 3 x 3

So, 990 = 2 x 5 x 3 x 3 x 11

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