# How to solve this monster?

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:

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

• no integer power of $10$ is an integer divisible by $17$ May 20 '17 at 2:49
• 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) May 20 '17 at 2:51
• 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). 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 $17 \mid 2^x \cdot 5^x$?