The $2019$th strictly positive integer $n$ such that $\binom{2n}{n}$ is not divisible by $5$

A possible way to find the answer:

Exploring the numbers for which $\left\lfloor \frac{2n}{5} \right\rfloor - 2 \left\lfloor \frac{n}{5} \right\rfloor =0$ shows the $3$ out of every $5$ integers satisfy this, with the $3$th, $6$th etc. cases being multiples of $5$.

Similarly exploring the numbers for which $\left\lfloor \frac{2n}{5^2} \right\rfloor - 2 \left\lfloor \frac{n}{5^2} \right\rfloor =0$ shows that $13$ out of every $25$ integers satisfy this, with the $13$th, $26$th etc. cases being multiples of $25=5^2$. Overlapping this with the previous result, $9$ out of every $25$ integers satisfy these two.

You can then prove that $\left\lfloor \frac{2n}{5^k} \right\rfloor - 2 \left\lfloor \frac{n}{5^k} \right\rfloor =0$ if and only if $n \in \{0,1,2,\ldots,\frac{5^k-1}{2}\} \mod 5^k$

and by induction that $\sum\limits_{k = 1}^t \left\lfloor \frac{2n}{5^k} \right\rfloor - 2\sum\limits_{k = 1}^t \left\lfloor \frac{n}{5^k} \right\rfloor = 0 $ for $3^t$ out of every $5^t$ integers with the $3^t$th, $2\times 3^t$th etc. being multiples of $5^t$.

To find the solution using this information:

• we can find $3^6=729\le 2019$ and $3^7=2187>2019$. So $t=7$ is too big.

• $2019 = 2\times 729+2\times243 + 2\times 27 + 2\times 9 +1\times 3$ which is $2\times 3^6+2\times3^5 + 2\times 3^3 + 2\times 3^2 +1\times 3^1$.

• The desired $n=2\times 5^6+2\times5^5 + 2\times 5^3 + 2\times 5^2 +1\times 5 ^1 = 37805$