I want to find $x$:


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:

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

  • 6
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $17 \mid 2^x \cdot 5^x$?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.