The number $25!$ has exactly 7 trailing zeros, true or false?

Question

I don't know how to determine it...

any hints?

Answer 1

Hint: a zero at the end comes from the product of a $2$ and a $5$. Since the $5$'s are more rare than the $2$'s, count how many $5$'s divide the numbers $1$, $2$, ..., $25$.

Here is the result, if you want to check your work:

$25!=15511210043330985984000000$

See also Trailing zero.

Answer 2

As others have noted, what counts is number of factors that are equal to $5$ in the number. Assuming you don't want to count the number of fives by hand, note that for $n!$ it is given by the formula: $$\sum_{k=1}^{\log_5 n}\big\lfloor n / 5^k\big\rfloor$$ Since one out of $5$ numbers has one factor, one out of $25$ adds another, etc.. By way of example, although $962!$ is an enormous number, you can easily calculate that it has exactly $238$ trailing zeroes, since: $$\bigg\lfloor \frac{962}{5}\bigg\rfloor+\bigg\lfloor \frac{962}{25}\bigg\rfloor+\bigg\lfloor \frac{962}{125}\bigg\rfloor+\bigg\lfloor \frac{962}{625}\bigg\rfloor$$ $$=192+38+7+1=238$$