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

sharan kumar
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    $168$ cannot be right because it is not a multiple of $20$. –  Aug 31 '19 at 20:37
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    Also a duplicate of this and friends. – Jyrki Lahtonen Aug 31 '19 at 20:40
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    Most be a typo. As Gae S say however many $0$s $38!$ ends with (so that $38! = M10^k$ and $10\not\mid M$) then $(38!)^{20}$ will have $20$ times as many. $(38!)^{20}=M^{20}10^{20k}$. And that can't be $168$. ...As there are $7$ multiples of $5$ up to $38$ and one of them is $25$ we have $8$ is the highest power of $5$ to divide $38!$ and as there are more multiples of $2$s than of $5$s, $8$ is the highest power of $10$ and $20*8 =160$. You are doing it right and the book has a typo. – fleablood Aug 31 '19 at 21:00
  • It would improve your Question to include the full problem statement in the body of the Question, not only in the title. – hardmath Sep 01 '19 at 00:09

2 Answers2

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$160$ 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, $8 \times 20 = 160$.

Sunghee Yun
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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.

Gnumbertester
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