Prove that there exists a multiple of $2016$, such that the multiple has only $4$ and $6$ as its digits.

Question

Prove that there exists a multiple of $2016$, such that the multiple has only $4$ and $6$ as its digits. Any ideas how to go about proving this? I tried it with pigeonhole-principle, i tried to find a similar demonstration like this one : Existance of multiple of $n$ with only 0 and 1 as it's digits , but i couldn't.

Answer 1

Use the pigeonhole principle to find a multiple of $63$ that is a repunit in base 100.

Multiply that number by $64$.

Note that $63\cdot 64=2\cdot 2016$.