Factorised factorial
Weekly Problem 17 - 2010
The value of the factorial $n!$ is written in a different way. Can you work what $n$ must be?
The value of the factorial $n!$ is written in a different way. Can you work what $n$ must be?
Age
14 to 16
Challenge level
Problem
For a positive integer $n$, we define $n!$ to be the product of all the positive integers from $1$ to $n$; that is $n!=1\times 2\times 3\times\ldots\times n$.
If $n!=2^{15}\times 3^6\times 5^3\times 7^2\times 11\times 13$, what is the value of $n$?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
Student Solutions
Answer: $n=16$
$13$ is the largest prime factor so $n$ is between $13$ and $17$
$7$ twice so one must come from $7$, one from $14$ so $n$ is at least $14$
$5$ three times so $n$ is at least $15$
$2$ fifteen times
Multiples of $2:$
$2,4$(two)$,6,8$(three)$,10,12,$(two)$,14$ makes eleven factors of $2$
So the number $n$ is $16$ (which itself is $2^4$)