Could you please tell me how many zeroes are there at the end of $36!$ to the power $36!$, i.e., $36!^{36!}$? I have been trying to find out. Read some reviews and answers related this but didn't understand at all. Thanks for the help.
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1The number of trailing zeroes of 36! equals the number of $5$'s that divide $36!$, which can be computed with De Polignac's formula. – vadim123 Apr 29 '15 at 03:02
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1Been wondering the same a week ago, hope this link helps! http://www.purplemath.com/modules/factzero.htm – Jesse P Francis Apr 29 '15 at 03:05
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1On a sarcastic note, if you just keep the number written as $36!^{36!}$, then the answer is zero. – Mike Pierce Apr 29 '15 at 03:21
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possible duplicate of How many zeroes are in 100! – apnorton Apr 29 '15 at 03:41
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@apnorton, not that simple! Its its a impossible number to the power an impossible number, making it further impossible! – Jesse P Francis Apr 29 '15 at 03:46
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@JessePFrancis Oops. I meant to delete that comment, because there's a slight difference ("in 100!" vs "at the end of 100!"). Other than that, however, the procedure is exactly the same. (Also, be careful saying a problem is "impossible." This problem is certainly solvable.) – apnorton Apr 29 '15 at 03:47
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It was simpler than I thought! :D – Jesse P Francis Apr 29 '15 at 04:12
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@mapierce271 looks at $36!^{36!}$ I see zero zeroes. – user3330644 Apr 29 '15 at 04:21
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De Polignac's formula tells us that $36!$ there are 34 powers of $2$ and 8 powers of $5$, so when $36!$ is factored, there are $8$ zeros because there is a $10^8$ in its factorisation. Since each power of $36!$ adds 8 zeroes, $36!^{36!}$ has $8\cdot36!$ zeros at the end.
Pauly B
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$\sum\limits_{k:5^k<n}\lfloor\frac{n}{5^k}\rfloor$ gives the number of trailing zeros in n!
Above suggests that there are eight trailing zeros in 36!, so we can write $36!=\text{some number ending with non zero digit, (say k)}\times 10^8$
$\implies 36!^{36!}=(k\times 10^8)^{36!}$
We just need to be concerned about the ${10^8}^{36!}$ part since $k^{36!}$ will not contribute any more zeroes (if last digit of k$\ne0$ then last digit of $k^m\ne0\forall m\in\mathbb{N}$
And hence it has $8.36!$ trailing zeros.
Jesse P Francis
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