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=\frac{2\cdot3\cdot\overbrace{(2\cdot2)}^4\cdot5\cdot\overbrace{(2\cdot3)}^6\cdot7\cdot\overbrace{(2\cdot4)}^8}{7\cdot4}=(8!)/(28)`

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


`k=n!=1440m\iff1440=\frac{n!}m`

which suggests that
`n=8`.