# 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 $$n=2\times 5^6+2\times5^5 + 2\times 5^3 + 2\times 5^2 +1\times 5 ^1 = 37805$$