<s>[INST] How to find the last non-zero digit of $50!$ A week ago i made a similar question but nobody help me, i´ve been trying but i still don't get it.
I want to know how to find the last non-zero digit of $50!$.
my try: 
First i have to know how much Zeros $50!$ has so i did this:
$$E_{5}50 = \sum _{1\leq k <n} \Bigg[\frac{50}{5^{k}}\Bigg] =\Bigg[\frac{50}{5}\Bigg] + \Bigg[\frac{50 }{25}\Bigg] = 12$$
So $50!$ has $12$ zeros which means that the last digit of $\quad\frac{50!}{10^{12}}\quad$ is the number that i´m looking for.
so if $x = \frac{50!}{10^{12}}$ i need to find $x (mod 10)$ to get it but this is such a big number and i still don't know how to reduce it.
Thanks in advance!!
 [/INST] My Solution:: calculation of last non zero digit in $50!$
Now first we will calculate number of zeros at the end of $50!$, which is obtained
when we multiplied $2$ and $5$. So we will calculate no. of $2$ and $5$ in $50!$
$\displaystyle 50! = \lfloor \frac{50}{2}\rfloor +\lfloor \frac{50}{2^2} \rfloor+\lfloor \frac{50}{2^3}\rfloor+\lfloor \frac{50}{2^4}\rfloor+\lfloor \frac{50}{2^5}\rfloor+........... = 25+12+6+3+1 = 47$
$\displaystyle 50! = \lfloor \frac{50}{5}\rfloor+\lfloor \frac{50}{5^2}\rfloor+..... = 10+2 = 12$
So Total no. of zeros at the end of $50! = 12$
Now $\displaystyle 50! = \lfloor \frac{50}{3}\rfloor+\lfloor\frac{50}{3^2}\rfloor+\frac{50}{3^3}\rfloor+..... = 16+5+1 = 22$
Now we can write $\displaystyle 50! = 1\times 3^{22}\times 7^8\times 11^4 \times 13^3 \times 17 \times 19\times  23\times 29\times 31 \times  17\times 37 \times 41 \times 43 \times 23 \times 47 \times 10^{12}\times 2^{35} = \bf{last digit\; (6)} = 6$
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