Find $144^3$ mod $213$
I'm not sure how to solve this.
I know that $213=3\times 71$, which are primes.
And I can find that $144\equiv 0$ mod $3$, and $144\equiv 2$ mod $71$.
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1Are you familiar with the Chinese remainder theorem? – saulspatz Mar 17 '20 at 22:27
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@saulspatz Yes. Do I want a solution which is $144^3$ mod $3$ and $144^3$ mod $71$ – AColoredReptile Mar 17 '20 at 22:28
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1Yes, that's right. – saulspatz Mar 17 '20 at 22:39
2 Answers
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$144^3\equiv0\pmod3$ and $144^3\equiv2^3=8\pmod{71}$.
If $x\equiv8\pmod{71},$ then $x\equiv79, 150, $ or $221\pmod{213}$.
Can you take it from here?
J. W. Tanner
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$144^{\large 3}\bmod\, 3\cdot 71\, =\, 3\:\!\underbrace{\left[\dfrac{\color{#0a0}{144^{\large 3}}}3\!\bmod 71\right]\!=\, 3\:\![\color{#c00}{50}]}_{\textstyle \ \dfrac{\color{#0a0}{2^{\large 3}}}3\,\equiv\, \dfrac{-63}3\,\equiv\,\color{#c00}{-21}}$
Bill Dubuque
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using $\ ab\bmod ac = a(b\bmod c) = $ Mod Distributive Law, and trivial mental arithmetic $\ \ \ $ – Bill Dubuque Mar 17 '20 at 22:45