If n is a positive integer and the product for all integers 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
OA =11
A.10
B.11
C.12
D.13
E.14
OA =11
990 = 9*10*11
Here the largest prime factor is 11.
To find any multiple of 990, it should have all the factors of 990.
The question also says what is the least value of n!
Therefore 11! is the answer because 11! = 11*10*9*8*........*2*1
11! will at least have all the factors of 990.
Hence B.
