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I want to find $x$:

$$10^x\,\text{mod}\,17=0$$

Here, even wolframalpha refuses to answer. If there may be no solution, then how to prove that there is no solution?

EDIT: The equation below is equivalent:

$$10^x-17\Big\lfloor \frac{10^x}{17}\Big\rfloor=0$$

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    $\begingroup$ no integer power of $10$ is an integer divisible by $17$ $\endgroup$
    – Will Jagy
    May 20 '17 at 2:49
  • $\begingroup$ If you could write a proof as an answer I would be happy @WillJagy (And by the way I think you commented my question/answer once because I know your nickname :D) $\endgroup$
    – KKZiomek
    May 20 '17 at 2:51
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    $\begingroup$ It might be of interest to you to work out the period of the powers $x$ which causes $10^x \bmod 17$ to repeat (never equaling zero, of course). $\endgroup$
    – hardmath
    May 20 '17 at 2:52
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Since $10 = 2 \cdot 5$, then $10^x = 2^x \cdot 5^x$. Since 17 is prime, is it ever going to be the case that $17 \mid 2^x \cdot 5^x$?

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  • $\begingroup$ I didn't think of that :D $\endgroup$
    – KKZiomek
    May 20 '17 at 2:56
  • $\begingroup$ @KKZiomek when doing modular arithmetic, it always helps to think back to the basics of divisibility. $\endgroup$
    – Oiler
    May 20 '17 at 2:59

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