Number of Trailing Zeros of Binomial Coefficient

Question

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 $x$.

Answer 1

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.