A repunit is a number that contains only “ones” (for example $111$, $1111111$,….). Prove that one can find a repunit divisible by $1973$

Question

It is a pigeonhole problem.

I have already known that there are $1972$ remainders in total and the two numbers which have the same remainder can be subtracted and the difference between the two numbers is divisible by $1973$.

BUT the difference is not a repunit number, it will be like $1111111...0000000$

Can someone help me out?

Thanks a lot!

Answer 1

Dividing that difference by $10^n$ for a suitable value of $n$ will not change whether it's divisible by $1973$. And thus you are done.

Answer 2

As an alternative solution, $1973$ is prime, so $10^{1972}\equiv1\mod1973$ by Fermat's little theorem,

so $1973$ divides $ \dfrac{10^{1972}-1}9.$