Number of zeros at the end of $10^{2}!+ 11^{2}!+12^{2}! \cdots+99^{2}!$

Question

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 $24$

Answer 1

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 $10^2!$.

Answer 2

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.