n is a positive integer, and k is the product of all integer

This is essentially a direct copy of an official question, with one number changed:

https://gmatclub.com/forum/if-n-is-a-po ... 90855.html

though you might notice how much more elegant the wording of the official problem is (there's no need to introduce two separate letters in this kind of problem).

Here, we know n! is divisible by 1440 = (12)^2 * (10) = (2^5)(3^2)(5). So we need n to be large enough so that we have one 5, two 3s, and five 2s among the divisors of the integers from 1 through n. It might be clear if you've done similar kinds of problems, or looked at prime factorizations of factorials, that we won't need an especially large value of n here, so you could just confirm that the smallest answer choice, n=8, works and be done here. But you could also solve without answer choices - as long as n is 5 or greater, we'll have a '5' in the product of n!, and as long as n is 6 or greater, we'll a 3 and a 6 in the product making up n!, so 3^2 will be a divisor of n!. Lastly, we just need to be sure to get five 2s. If n = 6, then n! = 6! is only divisible by 2^4, because we only get twos from 2, 4, and 6. But as long as n is 8 or greater, then 2^5 will easily divide n! (in fact, 2^7 will), so 8 is the smallest possible value of n.

BTGmoderatorDC wrote:n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is

A. 8
B. 12
C. 16
D. 18
E. 24

OA A

Source: Magoosh

Let’s break 1440 into prime factors:

1440 = 144 x 10 = 12 x 12 x 10 = 2^5 x 3^2 x 5^1

Thus, k/(2^5 x 3^2 x 5^1) = integer.

We also know that k is the product of all integers from 1 to n, inclusive, or, in other words, k = n!.

Let’s check our answer choices:

A. If n = 8, then k = 8! and 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 = 2^3 x 7 x 3 x 2 x 5 x 2^2 x 3 x 2 = 7 x 5 x 3^2 x 2^7 does contain five 2s, two 3s and one 5.

Answer: A