How many $0$'s are at the end of $(38!)^{20}$?

Question

I am getting $160$ as my answer but in the book, it is $168$. Which is the correct answer?

Answer 1

Your answer of $160$ is correct.

This is because $38!$ has $8$ trailing zeros. Take this to the twentieth power and you end up with $20\times8=160$ trailing zeros.

Answer 2

$160$ is the correct answer!

Proof) $$ \newcommand{\floor}[2]{\left\lfloor\frac{#1}{#2}\right\rfloor} \floor{38}{5} +\floor{38}{5^2} = 7 + 1 $$

Hence, $8 \times 20 = 160$.