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How do I find the number of zeros at the end of the Integer $$10^{2}!+ 11^{2}!+12^{2}! \cdots+99^{2}!$$

Answer provided for this question is $24$

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  • $\begingroup$ Can you answer a simpler question: how many zeros are at the end of the integer $100!$? $\endgroup$
    – Joey Zou
    Jul 16, 2016 at 8:19
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    $\begingroup$ HINT: This number is divisible by $10^2!$ but not by $10\cdot10^2!$. $\endgroup$ Jul 16, 2016 at 8:20

2 Answers 2

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The number of zeroes at the end of all numbers $10^{2}!,\ \ 11^{2}!,\ \ 12^{2}!,\ \ \cdots,\ \ 99^{2}!$, are equal or more than the one for $10^{2}!$.

So, you just need to count the number of zeroes at the end of $10^{2}!$

Moreover, we know that the number of zeroes at the end of $100!$ is equal to $\color{red}{\lfloor {\frac{100}{5}} \rfloor+\lfloor {\frac{100}{25}} \rfloor =24}$.

Revise: As Hagen mentioned here, the number of zeroes at the end of all the numbers $11^{2}!,\ \ 12^{2}!,\ \ \cdots,$ and $99^{2}!$ are more than $24$, that is the one for $10^2!$, and so the number of zeroes at the end of summation is equal to the number of zeroes at the end of $10^2!$.

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  • $\begingroup$ @Ayushakj: This answer, unlike Hagen von Eitzen’s, is incomplete. If some of the terms $n^2!$ with $n>100$ also ended in $24$ zeroes, it would in principle be possible for the sum to end in more than $24$ zeroes: the non-zero digits in the $25$-th place from the right could sum to a multiple of $10$. Hagen shows that this does not happen. $\endgroup$ Jul 16, 2016 at 20:35
  • $\begingroup$ I highly appreciate your attention to my answer. Some revisions have now been made to complete the solution. $\endgroup$
    – Majid
    Jul 17, 2016 at 3:32
  • $\begingroup$ Looks fine now; +1. $\endgroup$ Jul 17, 2016 at 3:33
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    $\begingroup$ @Axoren when there are at least 25 zeroes at the end of the others, it does not matter what happens during the summation, because they finally sum up with $10^2!$ which has only $24$ zeroes at the end, and so the number of zeroes at the end of the total summation is $24$. $\endgroup$
    – Majid
    Jul 17, 2016 at 5:29
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    $\begingroup$ @Axoren $121!$ has $28$ zeroes at the end which is strictly more than $24$ as the number of zeroes at the end of $100!$. $\endgroup$
    – Majid
    Jul 17, 2016 at 8:02
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The number of zeroes at the end of $n!$ is $\lfloor \frac n5\rfloor+\lfloor \frac n{25}\rfloor+\lfloor \frac n{125}\rfloor+\ldots$. In particular, $10^2!$ ends in $\lfloor\frac{100}5\rfloor +\lfloor\frac{100}{25}\rfloor+0=24$ zeroes. On the other hand, for $n\ge 105$ (e.g., for $n=11^2,\ldots, 99^2$), the number $n!$ ends in at least $\lfloor\frac{105}5\rfloor +\lfloor\frac{105}{25}\rfloor=25$ zeroes. Adding such a number to $10^2!$ does not change any of the lower 25 digits and hence does not change the numbre of zeroes at the end.

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  • $\begingroup$ How to you know the number of zeroes at the end of $n!$ is $\lfloor \frac n5\rfloor+\lfloor \frac n{25}\rfloor+\lfloor \frac n{125}\rfloor+\ldots$? $\endgroup$ May 31, 2017 at 17:40

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