Zeroes at the end of an expression.

Question

Please help me with this problem

How many zeroes are there at the end of the expression $4!^{4!}+8!^{8!}+16!^{16!}$

Answer 1

$8!$ has one zero at its end as $5,2<8$. So there are $8!$ zeros at the end of $(8!)^{8!}$. $5,10,15<16$, so there are three zeros at the end of $16!$. So there are $3\cdot 16!$ zeros at the end of $(16!)^{16!}$. However $(4!)^{4!}$ has no zeros at the end of it. Hence the sum $(4!)^{4!}+(8!)^{8!}+(16!)^{16!}$ has no zeros at the end of it. The last digit of this number is same as the last digit of $24^{24}$, which is $6$.

Answer 2

Note that $8!$ and $16!$ are divisible by $5$, but $4!$ is not.