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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!

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  • $\begingroup$ What do you mean by finding them? If you want an explicit formula, I don't think that is possible. $\endgroup$ – Emre Jan 17 '17 at 0:56
  • $\begingroup$ Actually, I need only to show that there are exactly two non-trivial solutions. $\endgroup$ – nag Jan 17 '17 at 0:57
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    $\begingroup$ What counts as "trivial" solutions? $x \equiv 0,1$ mod $10^k$? $\endgroup$ – Torsten Schoeneberg Jan 17 '17 at 0:59
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    $\begingroup$ Yes, these ones $\endgroup$ – nag Jan 17 '17 at 1:00
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    $\begingroup$ Since two answers have already been posted saying that $x(x-1) \equiv 0 \pmod{10^k}$ implies either $x\equiv 0 \pmod{10^k}$ or $x-1\equiv 0 \pmod{10^k}$, and were subsequently deleted after a simple counterexample ($x=6,k=1$) was pointed out, I thought I'd make note of those events in a comment. I hope this saves anyone else the wasted effort of posting such an answer. $\endgroup$ – David K Jan 17 '17 at 1:10
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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.
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  • $\begingroup$ >.< And this is what happens when you don't learn in any serious kind of way. $\endgroup$ – Simply Beautiful Art Jan 17 '17 at 1:02
  • $\begingroup$ Oh yes, how could I forgot about CRT $\endgroup$ – nag Jan 17 '17 at 1:03
  • $\begingroup$ @SimpleArt I don't understand what you're trying to say. $\endgroup$ – Emre Jan 17 '17 at 1:05
  • $\begingroup$ Oh, its just that I don't spend time actually trying to learn math. I just pick up bits and pieces and put them together (hopefully well), which is why I can solve some integrals with complex analysis techniques but lack a basic understanding of real analysis. $\endgroup$ – Simply Beautiful Art Jan 17 '17 at 1:07

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