If n is a positive integer and product of all the 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
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 = (2)(2)(2)
(3)
70 is divisible by
5 <--> 70 = (2)
(5)(7)
330 is divisible by
6 <--> 330 =
(2)(3)(5)(11)
56 is divisible by
8 <--> 56 =
(2)(2)(2)(7)
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
