Number of Trailing Zeros of Binomial Coefficient If $x+2=18181818...$ $n$ $digits$, find the number zeros at the end of ${x \choose x/2}$. I have tried using Legendre's formula for factorials, but I have got nowhere because of the strange value of $x$.
 A: We want to find how many Trailing zeros for $\binom{2n}{n}$.
First Write the binomial as factorials $\frac{(2n)!}{(n!)^2}$
And apply Legendre formula for every term separately.
For $(2n)!$ there are $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{2^k} \rfloor $ $2$'s and  $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{5^k} \rfloor $ $5$'s
Now for $(n!)^2$ it have the same powers of primes as $n!$ but they are multiplied by $2$ (simple powers rule), so there are $\sum \limits_{k=1}^{\infty} 2\lfloor \frac{n}{2^k} \rfloor $ $2$'s and  $\sum \limits_{k=1}^{\infty} 2\lfloor \frac{n}{5^k} \rfloor $ $5$'s
So for the expression $\frac{(2n)!}{(n!)^2}$ there are $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{2^k} \rfloor-2 \lfloor \frac{n}{2^k} \rfloor $ $2$'s and  $\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{5^k} \rfloor -2\lfloor \frac{n}{5^k} \rfloor$ $5$'s (powers in denominator are subtracted ,simple powers rule).
So $\frac{(2n)!}{(n!)^2} $ there are $Min(\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{2^k} \rfloor-2 \lfloor \frac{n}{2^k} \rfloor ,\sum \limits_{k=1}^{\infty} \lfloor \frac{2n}{5^k} \rfloor-2 \lfloor \frac{n}{5^k} \rfloor )$ Trailing Zeros,
Checked for first $1000$ numbers and its correct.
