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

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.

$$29\equiv7,27\equiv5\pmod{11}$$

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

$$\implies7x^3\equiv5\pmod{11}$$

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

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

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

$$x=x^{10}(x^3)^{-3}\equiv1\cdot(-4)^{-3}\equiv-2^{-6}\equiv-2^4\equiv6$$

• $3=10−3⋅3$. should be $1=10−3⋅3$ Jun 20 '20 at 12:08
• As lab bhattacharjee hasn't responded so far, I've corrected the post. Jun 20 '20 at 12:22
• @TobyMak, Thanks for the edit Jun 20 '20 at 12:34
• Posted five minutes after the question. Do you ever search for duplicates? This type of equations have been covered many times already. Jun 20 '20 at 19:02
• @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. 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 $$x=6$$.