How many is different powers of 10, which divide 1000! ? Powers are supposed to be positive integers
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$50!$ or $1000!$? – hmakholm left over Monica Oct 03 '15 at 12:21
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4See also number of zeroes at the end of factorial. – hmakholm left over Monica Oct 03 '15 at 12:23
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oh, my bad. 1000! of course – EnemyPanda Oct 03 '15 at 12:24
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So why did you write 50! in the title? – JRN Oct 03 '15 at 13:17
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1@Scientifica Please don't make relatively trivial edits directly in the Low Quality and Close queues, as it will unilaterally complete the review. From the description: "Edit if you can fix all the problems with this [post]". – epimorphic Oct 04 '15 at 16:37
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@epimorphic Yes you're right. If ever I want to edit without fixing all the problems, I should skip or recommend close the question before editing it. Most of the time I forget after editing I can't flag the question anymore (especially if it's for using LaTeX). I apologize and thank you for your important comment! – Scientifica Oct 04 '15 at 20:44
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In this answer, it is shown that the number of factors of the prime $p$ in $n!$ is $$ \frac{n-\sigma_p(n)}{p-1} $$ where $\sigma_p(n)$ is the sum of the digits in the base-$p$ representation of $n$.
Since $1000=13000_{\text{five}}$, the number of factors of $5$ in $1000!$ is $$ \frac{1000-4}{5-1}=249 $$ Since $1000=1111101000_{\text{two}}$, the number of factors of $2$ in $1000!$ is $$ \frac{1000-6}{2-1}=994 $$ Thus, $10^{249}\mid1000!$ Therefore, the positive powers of $10$ that divide $1000!$ are $$ 10^1, 10^2,10^3,\dots,10^{249} $$
