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Sorry, I'm not into advanced math, but it wonders me, why factorials above ~85! contain lots of zero's at the end.

Example, 100! = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

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    $\begingroup$ 85 is no benchmark. The higher you get the more 0's you will have. $\endgroup$ – user88595 Feb 10 '14 at 15:25
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    $\begingroup$ Yes, the question is why. $\endgroup$ – niklon Feb 10 '14 at 15:25
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    $\begingroup$ Solution below explains it well. It's all about powers of 2 and 5. $\endgroup$ – user88595 Feb 10 '14 at 15:26
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    $\begingroup$ Why'd you pick 85? Were the 19 zeros at the end of 84! below the "lots" cutoff? $\endgroup$ – user2357112 Feb 11 '14 at 3:01
  • $\begingroup$ Honestly, I picked it by accident. Because lower factorials didn't concern me that much, so I chose 85 as a milestone. $\endgroup$ – niklon Feb 11 '14 at 12:00
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A number ends in $m$ zeroes if and only if it is divisible by $10^m$. To be divisible by $10^m$ means to be divisible by $2^m$ and by $5^m$. Big factorials (nothing special about 85) are products of lots and lots of numbers. After a while, lots of those numbers are divisible by 2 (and powers of 2), and lots are divisible by 5 (and powers of 5), so the product is divisible by a really high power of 10. (For more precise mathematical statements about how many zeroes you'll get, search this site, where there are many variations on this question.)

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It should be pretty obvious why $10!$ and all higher factorial must all have at least one zero at the end: they're all divisible by $10$.

$$10! = \mathbf{10} \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$

If you think about it a bit more, it's also pretty obvious that $20!$ and any factorials above it must haveat least two zeros at the end (because they're divisible by $10 \times 20$) and that $30!$ and above must have at least three zeros (because they're divisible by $10 \times 20 \times 30$) and so on.

Actually, this "rule" underestimates the number of zeros at the end of factorials by about a factor of $2$. Why? Because $2 \times 5 = 10$, so $5! = 5 \times 4 \times 3 \times 2 \times 1$ already has one zero at the end, and every further multiple of $5$ adds yet another zero (there being plenty of even numbers to provide the multiples of $2$ needed to make up $10$). So $15!$ has three zeros at the end, not just one, and $20!$ actually has four, not two.

Also, $25!$ actually gains two extra zeros from being a multiple of $25 = 5 \times 5$, for a total of six. The same happens at $50!$ and $100!$, and $125!$ actually has three more zeros at the end than $124!$, because $125 = 5 \times 5 \times 5$.

So, looking at your example, we don't actually need to calculate $85!$ to tell that it has twenty zeros at the end: one each from $5$, $10$, $15$, $20$, $25$, $30$, $35$, $40$, $45$, $50$, $55$, $60$, $65$, $70$, $75$, $80$ and $85$, and one extra each from $25$, $50$ and $75$.

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First of all, you need to know that starting with $n = 5$, $n!$ will always have at least one zero in the end.
$5! = 5\times4\times3\times2\times1 = 120$. In this example, the $5$ and one $2$ contribute to that zero. After this, at $10!= 3628800$, another zero is added because the $5$ from the '10' and another $2$ from the product $10!$ contribute to the extra zero. Since the number of '2's in $n!$ will always be greater than number of $5$s, the number of zeroes at the end of $n!$ will be equal to the total number of 5s in the product $n!$.

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