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I want to find The $25$th digit of $100!$.

My attempt:It is easy to know it has $24$ zeroes.Because:

$\lfloor {\frac{100}{5}} \rfloor+\lfloor {\frac{100}{25}} \rfloor =24$

By getting the fist digits(after deleting all $5$ factors and $24$,$2$ factors)and multiplying them to each other we get the answer $4$ but I want an easier way.

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Observe the 24 zeros, divide them out, then

$$ x \equiv \frac{4^{10} 9^{10} 20!}{5^4} \equiv 6^2 {4!}^4 \pmod{10} $$

Obviously $ x \not \equiv \{5, 0\} \pmod {10} $, due to $ 5 \nmid x $. Similarly, it contains factor $ 2 $, so it must be either $ 2, 4, 6, 8 $. At this point, you can simple evaluate it, and you will get $ 4 $.

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  • $\begingroup$ Answer is 3 ( 100! is 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000, calculator used www.careerbless.com/calculators/ScientificCalculator/ ) $\endgroup$ – Kiran Jul 16 '16 at 10:36
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    $\begingroup$ Sorry for misinterpreting this. I thought it is from left. $\endgroup$ – Kiran Jul 16 '16 at 10:44
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    $\begingroup$ @M.Fischer Can we find 25th digit from right ?, What if there were 10 zeroes then can we still find the 25 digit from right ? $\endgroup$ – A---B Jul 16 '16 at 17:21
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    $\begingroup$ @ritwiksinha That's harder. The best way to do so would be to first find out the number of digits (i.e. reduce it to the form $ x = 10^n c $ with $ 0.1 < c < 1 $, and then find the $ n - 25 $ digit (divide $ 10^{n - 25} $ out under modulo 10) $\endgroup$ – user347499 Jul 17 '16 at 8:27
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    $\begingroup$ Prime factor $ 100! $ by grouping the divisors together. Let $ 100! = \prod_{n=1}^100 n $, then you can simply "partition" it into smaller products, $ (\prod_{n = 0, 5 \nmid 2n + 1}^{50} 2n + 1) (\prod_{n = 5, 5 | n}^{100} n)(\prod_{n = 1, 5 \nmid n, 2|n}^{100} n) $ (odd, even, products of fives, respectively). $\endgroup$ – user347499 Jul 24 '16 at 10:30
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If we exclude the numbers that are divisible by 5, we see a cycle that repeats.

$4! = 24\\ 9!/6!\equiv 24 \mod 100$

$100! = \frac {100!}{5^{20} 20!} (5^{20} 20!) = \frac {100!}{5^{20} 20!}\frac {20!}{5^4 4!} (4!)(5^{24})$

$(24^{25})(5^{24}) = (12^{25})(10^{24})(2)$

$12^{25} \equiv 12^5 \mod 100\equiv 32 \mod 100$

The last $2$ non-zero digits of 100! are $64.$

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