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Someone wrote that $625!$'s last $156$ digits are zeros because $125+25+5+1=156$. If it's true that $625!$ has $156$ zeros at the end, how does "$125+25+5+1=156$" prove it?

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  • $\begingroup$ The answer is here: math.stackexchange.com/questions/1384419/… $\endgroup$ Commented Oct 10, 2015 at 13:52
  • $\begingroup$ Why does the number of zeroes step up at each multiple of 5? $\endgroup$
    – user265554
    Commented Oct 10, 2015 at 13:54
  • $\begingroup$ It's a count of how many times $5$ appears as a factor in the product $625!$ : $125=\frac{625}{5}$ gives the number of multiples of $5$ less than $625$, $25=\frac{625}{25}$ gives the number of multiple of $25$ (who need to be counted a second time since they provide $2$ factor instead of $1$), etc. Since there must be at least as many factors $2$, it gives the highest power of $10$ that divides $625!$, i.e. the number of $0$'s. $\endgroup$
    – Arnaud D.
    Commented Oct 10, 2015 at 13:56
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    $\begingroup$ en.wikipedia.org/wiki/Legendre%27s_formula $\endgroup$
    – user236182
    Commented Oct 10, 2015 at 14:21

1 Answer 1

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When you factorize $625!$, you are interested in the number of $10$s you can divide it by until it is no longer possible to divide evenly.

But $10=2\cdot 5$, so you actually want to know how many $2$s and $5$s are in the prime factorization of $625!$. You will quickly realize if you start listing the product that there are many many more factors of $2$s than there are $5$s, which means that the number of zeroes at the end is exactly equal to the number of $5$s in the prime factorization of $625!$.

How many $5$s are there? Well every multiple of $5$ in the product will count as one five. But every multiple of $25$ will count as an extra $5$, and every multiple of $125$ will count an extra on top of that, and every multiple of $5^4 = 625$ will count a fourth time.

There are $125$ positive multiples of $5$ less than or equal to $625$, $25$ multiples of $25$ less than or equal to $625$, $5$ multiples of $125$ less than or equal to $625$, and just the one multiple of $625$.

This gives your $$125 + 25 + 5 + 1 = \boxed{156}$$

zeroes.

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  • $\begingroup$ Thank you! Would you happen to know any place (book or a web page etc.) where I could find more stuff like this? Every now and then I come across these theorems that quite easily tell you about properties of numbers, but I won't even know that these theorems exist unless I luckily stumble upon them. $\endgroup$
    – user265554
    Commented Oct 10, 2015 at 14:23
  • $\begingroup$ Here is a link to a free introductory number theory text. If you are a high school student and like math I strongly recommend looking into PROMYS. $\endgroup$ Commented Oct 10, 2015 at 14:27
  • $\begingroup$ The art of problem solving website has good forums where you can ask or read about any kind of math. $\endgroup$ Commented Oct 11, 2015 at 2:00

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