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Solve $x^{67}\equiv 5\pmod{101}$.

I used this How to solve $x^5\equiv 3 \pmod{19}$? to solve this problems, I would like to make a solution verfication.

I'll solve the equation $67k\equiv 1 \pmod{100}$.

Multiply by $3$ ,$201k\equiv 3 \pmod{100} \implies k\equiv 3 \pmod{100}\implies k=3$.

So $x=5^3\pmod{101}\equiv 24\pmod{101}$.

Is my solution correct ?

Thanks !

Algo
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  • Looks ok to me. For full credit you probably should explicitly mention that 101 is a prime :-) – Jyrki Lahtonen Jul 02 '22 at 17:00
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    Yes, it is correct, Please only ask SV questions if you have a *mathematical* question about a *specific* step in your proof. Which step do you doubt, and why? $ $ Note $\bmod 100\!:\ \dfrac{1}{67}\equiv \dfrac{1}{-33}\equiv \dfrac{1}{-33}\dfrac{3}3\equiv \dfrac{3}1\ $ by [Gauss's algorithm](https://math.stackexchange.com/a/174687/242). – Bill Dubuque Jul 02 '22 at 17:00
  • Remember, it is easy to check your result. You can do it by hand, or use something like [Wolfram Alpha](https://www.wolframalpha.com/input?i=24%5E67+mod+101). – lulu Jul 02 '22 at 17:03
  • See [this Linked list](https://math.stackexchange.com/questions/linked/3187793?lq=1) for *many* more worked examples. – Bill Dubuque Jul 02 '22 at 17:15

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