Non-trivial solutions of $x(x-1) \equiv 0 \ (\text{mod} \ 10^k)$

Question

I need to show that there are exactly two non-trivial solutions of $$x(x-1) \equiv 0 \ (\text{mod} \ 10^k)$$for any $k$.

The only thing I've figured out is that one of the numbers has to be divisible by $2^k$ and another one has to be divisible by $5^k$ (and it could not be even). I'd be happy if someone gives me any help or advice!

Answer 1

You're right about your observation, let's look at the cases:

1. If $2^k\lvert x$, then $5^k\lvert x-1$. So, $$x\equiv 0\pmod{2^k}\text{ and } x\equiv 1\pmod{5^k}$$ By Chinese Remainder Theorem, there is a unique $x$, between $0$ and $10^k-1$ satisfying these equations.
2. Similarly, if $2^k\lvert x-1$, then $5^k\lvert x$. So, $$x\equiv 1\pmod{2^k}\text{ and } x\equiv 0\pmod{5^k}$$ By Chinese Remainder Theorem, there is a unique $x$, between $0$ and $10^k-1$ satisfying these equations, too. And clearly two cases give different solutions.