I have been told to use Fermat-Euler theorem. Using it, I get to $2^{(323)(288)} \equiv 1 \pmod{323}$, but I am not sure where to go from there!
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do you mean $$2^{300} \mod 323$$? – Dr. Sonnhard Graubner Mar 20 '17 at 12:50
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Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Mar 20 '17 at 12:53
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See this thread for a compendium of techniques needed to do problems like this. IMO this is a duplicate, but I have promised not to vote to close "abstract duplicates" given that my vote would be immediately binding. – Jyrki Lahtonen Mar 20 '17 at 14:18
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We know $\varphi(323)=288$. This means $$2^{288}\equiv 1\mod 323$$ Thus, $$2^{300}\equiv2^{300-288}\equiv2^{12}\mod 323$$ And $2^{12}=4096$, so from this point on, there's no large numbers to worry about.
To finish the calculation simply note that $2^{12}\equiv 4096\equiv 220\mod 323$, and thus
$$2^{300}\equiv 220\mod 323$$
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Method$\#1:$
As $323=17\cdot19$
$2^4\equiv-1\pmod{17}\implies2^8\equiv1$ as $300\equiv4\pmod8,2^{300}\equiv2^4\pmod{17}\equiv-1\ \ \ \ (1)$
$2^9\equiv-1\pmod{19}\implies2^{18}\equiv1,300\equiv-6\pmod{18},$
$2^{300}\equiv2^{-6}\pmod{19}\equiv7^{-1}\equiv11\ \ \ \ (2)$ as $7\cdot11\equiv1\pmod{19}$
Apply CRT on $(1),(2)$
Method$\#2:$
$\lambda(323)=$lcm$(17-1,19-1)=144$
$300\equiv12\pmod{144}\implies2^{300}\equiv2^{12}\pmod{323}\equiv?$
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