Find all positive integers $x$, that satisfy $29x^{33} \equiv 27 \pmod {11}$.

I approached this the following way:

Since from $29x^{33} \equiv 27 \pmod {11}$ we get that $7x^{33} \equiv 5 \pmod {11}$ and since $\gcd(7,5)=1$ we would get that $\phi(11)=10$ which would imply that $7x^{10} \equiv 5 \pmod {11}$.

How should I continue from here, it doesn't seem to be quite clear.



and $$\phi(11)=10$$


Now as $7\cdot8\equiv1\pmod{11},$

$$ x^3\equiv5\cdot8\equiv(-4)\pmod{11}$$

Finally as $1=10-3\cdot3,$


  • 3
    $\begingroup$ $3=10−3⋅3$. should be $1=10−3⋅3$ $\endgroup$ Jun 20 '20 at 12:08
  • $\begingroup$ As lab bhattacharjee hasn't responded so far, I've corrected the post. $\endgroup$
    – Toby Mak
    Jun 20 '20 at 12:22
  • $\begingroup$ @TobyMak, Thanks for the edit $\endgroup$ Jun 20 '20 at 12:34
  • $\begingroup$ Posted five minutes after the question. Do you ever search for duplicates? This type of equations have been covered many times already. $\endgroup$ Jun 20 '20 at 19:02
  • $\begingroup$ @Jyrki, Please feel free to mark it as duplicate. Also, in MSE, this is not the only type of questions which have been asked many a times. $\endgroup$ Jun 21 '20 at 1:27

as lab showed : $$x^3\equiv-4$$ then we have: $${(x^3)}^7\equiv{(-4)}^7$$

$$ \Longrightarrow x \equiv6 \,(mod11) $$


Another path: Subtract $11$ from $29$ to get

$$18x^3\equiv 27 \pmod{11}.$$

Divide by $9$

$$2x^3 \equiv 3 \equiv 14 \pmod{11}.$$

Divide by $2$

$$x^3 \equiv 7 \pmod{11}.$$

Cube both sides

$$x^9 \equiv x^{-1} \equiv 49\cdot 7 \equiv 2 \pmod{11}.$$

Solve $2x\equiv 1 \pmod{11}$ to get $x=6$.


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