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
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.
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.
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