What is the method of solving similar problems like given below $$x^7 \equiv 25\pmod{54}$$
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1Take $x=43$ this is the solution – Dr. Sonnhard Graubner Dec 06 '19 at 15:46
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1@Dr.SonnhardGraubner The question is not about what the solution is but about the method. – Arnaud Mortier Dec 06 '19 at 15:48
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1Hint: $x$ is coprime to $54$ so $x^{18}\equiv 1$ so expt arith is $!\bmod 18$ so you need only raise both side to power $,1/7\equiv -5\pmod{18},$ yielding $,x\equiv 25^{-5}\equiv 43,\ $ i.e take the $7$'th root of both sides by raising to power $,1/7.\ $ The question is a dupe and will likely be closed as such. – Bill Dubuque Dec 06 '19 at 15:49
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First solve $x^7\equiv25\bmod27$ as follows.
Note that $x^{18}\equiv1\bmod27$ by Euler's theorem, so
$x^7\equiv-2\bmod27\implies (x^7)^{13}\equiv x^{91}\equiv (x^{18})^5x\equiv x\equiv(-2)^{13}\bmod 27.$
Now $2^5=32\equiv5\bmod27$, so $2^{10}\equiv25\equiv-2\bmod27,$ so $x\equiv(-2)^{13}\equiv2\times8\equiv16\bmod 27.$
To have $x^7\equiv25\bmod 54$, we also need $x\equiv1 \bmod2$. Can you take it from here?
J. W. Tanner
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I also know $2^{10}\equiv25\bmod27$ because I know $27|999$ and $2^{10}=1024$ – J. W. Tanner Dec 06 '19 at 16:11
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I could have said $\bmod 27: x^7\equiv5^2\implies x\equiv(x^7)^{13}\equiv ((5)^2)^{13}\equiv5^{26}\equiv5^8\equiv(5^2)^4\equiv(-2)^4\equiv16$ – J. W. Tanner Dec 08 '19 at 00:28