What is the first non zero digit in 50 factorial (50!)? What is the first non zero digit in 50 factorial (50!)?
Any Help or hint will be appreciated.
 A: Since $50_{\text{ten}}=200_{\text{five}}$, $\sigma_5(50)=2$. Thus, the number of factors of $5$ in $50!$ is $\frac{50-2}{5-1}=12$.
Since $50_{\text{ten}}=110010_{\text{two}}$, $\sigma_2(50)=3$. Thus, the number of factors of $2$ in $50!$ is $\frac{50-3}{2-1}=47$.
Thus, $\frac{50!}{10^{12}}$ has $35$ factors of $2$.
Little Fermat says that $2^4\equiv1\pmod{5}$, so $2\cdot2^4\equiv2\pmod{10}$, therefore $2^{35}\equiv2\cdot2^{34}\equiv2\cdot2^2\equiv8\pmod{10}$.

Every integer is uniquely representable as $m\cdot2^j\cdot5^k$ where $(m,10)=1$.
Let's compute the product $\bmod{\,10}$ of all of the numbers $m$ so that $(m,10)=1$ and $m\cdot2^j\cdot5^k\le50$, grouped by $2^j\cdot5^k$:
$\overbrace{(1\cdot3\cdot7\cdot9)^5}^{1}\overbrace{(1\cdot3\cdot7\cdot9)^2(1\cdot3)}^{2}\overbrace{(1\cdot3\cdot7\cdot9)^1(1)}^{4}\overbrace{(1\cdot3)^{\vphantom{1}}}^{8}\overbrace{(1\cdot3)^{\vphantom{1}}}^{16}\overbrace{(1)^{\vphantom{1}}}^{32}\\
\overbrace{(1\cdot3\cdot7\cdot9)^1}^{5}\overbrace{(1\cdot3)^{\vphantom{1}}}^{10}\overbrace{(1)^{\vphantom{1}}}^{20}\overbrace{(1)^{\vphantom{1}}}^{40}\\
\overbrace{(1)^{\vphantom{1}}}^{25}\overbrace{(1)^{\vphantom{1}}}^{50}\\
\equiv(1\cdot3\cdot7\cdot9)^9(1\cdot3)^4\equiv9\pmod{10}$

Thus, $\frac{50!}{10^{12}}\equiv8\cdot9\equiv2\pmod{10}$.
A: First, the exponent of $2$ in $50!$ is $47$, and the exponent of $5$ in $50!$ is $12$, so we’re looking for $\frac{50!}{10^{12}}$ mod$ 10$.

Since $\frac{50!}{10^{12}}$ is even (in fact, divisible by $2^{35}$), it suffices to compute it mod $5$:
$\frac{50!}{5^{12}}$ $\equiv$ ($4!$. $1$. $4!$. $2$. $4!$. $3$. $4!$. $4$. $4!$. $1$.($4!$. $1$. $4!$. $2$. $4!$. $3$. $4!$. $4$. $4!$). $2$
=$4!^{10}$.$4!^2$.2!
$\equiv$$(-1)^{10}$.$(-1)^2$.2
$\equiv$ $2$ mod $5$
$\frac{50}{10^{12}}$ $\equiv$ $2$.$2^{-12}$ $\equiv$ $2$ . $(2^4)^{-3}$ $\equiv$ $2$ mod $5$

So, $\frac{50!}{5^{12}}$ $\equiv$ $2$ (mod $10$ ) 
A: The obvious method is to compute $50!$ and look at its digits (there's not that many of them).  Here it is:

30414093201713378043612608166064768844377641568960512000000000000

This is fairly straightforward to do on a computer (as long as you're alert for numerical overflow).  This method also has the benefits of (a) giving you all the other digits too, which helps you cross-check that the answer is correct, (b) being less likely to incur human error, and (c) will be more scaleable than human-operated methods.
In Wolfram|Alpha, we can input 50! to obtain:

I personally prefer to use GAP.  In GAP, we input Factorial(50); to obtain:
gap> Factorial(50);
30414093201713378043612608166064768844377641568960512000000000000

Or if we're feeling industrious, we might whip up some code to compute the last non-zero digit of $n!$ for all $n \in \{1,2,\ldots,100\}$.
for n in [1..100] do
  N:=Factorial(n);
  str:=DigitsNumber(N,10);
  i:=Size(str);
  while(str[i]='0') do
    i:=i-1;
  od;
  Print(n," ",Int([str[i]]),"\n");
od;

We can put these numbers into Sloane's On-Line Encyclopedia of Integer Sequences, and discover it's sequence A008904.  Here we can learn all sorts of things about these numbers, along with code for SAGE, Python, Mathematica and PARI (much more efficient than mine above).
