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 the least possible value of n?

A. 10B. 11C. 12D. 13E. 14

Option B.

Step-by-step explanation:

n is a positive integer and product of all integers from 1 to n is a multiple of 990.

That means n! = 990 * k

Or n! = 2 * 5 * 3 * 3 * 11 * k

If n = 2, then 2! will not be a multiple of 990.

If n = 3, then 3! will not be a multiple of 990.

If n = 5, then 5! will not be a multiple of 990.

If n = 9, then 9! will not be a multiple of 990 because 11 will be missing.

Now if we take n = 11, then 11! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 = 39916800 which is divisible by 990.

[As we can see all multiples of 990 = 2, 5, 9, 11 are present in 11! ]

Therefore, option B. 11 will be the answer.