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**melese** Multiplying the congruence $\displaystyle x^5\equiv6\pmod{17}$ by $\displaystyle 3$ we get, equivalently, $\displaystyle 3x^5\equiv18\equiv1\pmod{17}$.

There exists a positive integer $\displaystyle m $ such that $\displaystyle x\equiv3^m\pmod{17}$ and so $\displaystyle 3x^5\equiv3^{5m+1}\equiv{17}$. Finally, $\displaystyle 3^{5m+1}\equiv1\equiv3^{16}\pmod{17}$, which is equivalent to solving $\displaystyle 5m+1\equiv{16}\equiv0\pmod{16}$. The solution in the range $\displaystyle 0\leq i\leq16$ is $\displaystyle m=3$.

The solutions of the original congruence are given by $\displaystyle x\equiv3^5\equiv10\pmod{17}$.