Question

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

Answer 1

$k=n!=1\cdot2\cdot\cdots\cdot(n-1)\cdot n$

Prime factorization of 1440:

$1440=2^5\cdot3^2\cdot5$

The smallest *distinct* integer factors of 1440 can be pulled from the prime factorization:

$1440=\dfrac{2\cdot3\cdot\overbrace{(2\cdot2)}^4\cdot5\cdot\overbrace{(2\cdot3)}^6\cdot7\cdot\overbrace{(2\cdot4)}^8}{7\cdot4}=\dfrac{8!}{28}$

We're told that $k=1440m$ for some integer $m$. So

$k=n!=1440m\iff1440=\dfrac{n!}m$

which suggests that $n=8$.