DFritschy wrote: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) 10 B) 11 C) 12 D) 13 E) 14
Thanks!
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 = 2x2x2x
3
70 is divisible by
5 <--> 70 = 2x
5x7
330 is divisible by
6 <--> 330 =
2x
3x5x11
56 is divisible by
8 <--> 56 =
2x
2x
2x7
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 a 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
