If n is a positive integer and the product of all the integers from 1 to n, inclusive, is multiple of 990, what is the least value of n?
A. 10
B. 11
C. 12
D. 13
E. 14
A. 10
B. 11
C. 12
D. 13
E. 14
The product of all the integers 1 to n is n!. We need the smallest integer value of n such that n! will be divisible by 990.sugmomo wrote:If n is a positive integer and the product of all the integers from 1 to n, inclusive, is multiple of 990, what is the least value of n?
A. 10
B. 11
C. 12
D. 13
E. 14
990 = 10*99 = 2*5*9*11. Thus, n! must be divisible by 2, 5, 9 and 11.
If n=10, 10! will not be divisible by 11. Eliminate A.
If n=11, 11! will be divisible by 2, 5, 9, and 11. Thus, the smallest value of n that will work is n=11.
The correct answer is B.
