\(n\) is a positive integer, and \(k\) is the product of all integers from \(1\) to \(n\) inclusive. If \(k\) is a multi

VJesus12 wrote: ↑

Fri Dec 17, 2021 2:06 am

\(n\) 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

Answer: A

Source: Magoosh

The problem is saying that

K=N!

The smallest multiple of 1440 is 1440 and can be decomposed to

32*9*5, or 2^5*3^2*5

So in choosing N the multiplication down to 1 must include as a minimum
5 instances of 2 as a factor, 2 instances of 3 and 1 of 5.

Writing down some multiples of 2 to get started

2 4 6

These numbers contain a total of 4 instances of 2 as a factor. So we know we need 1 more. Therefore N has to be 8 or more to provide the 5 instances.


But is that enough ? Does 1 through 8 multiplied contain the two instances of 3 and 1 instance of 5?

1-8 contains a 3 and a 6, so two instances of 3 are covered. It also has one instance of 5, so the answer is

8, A