Determine the remainder, when $77^{200}$ is divided by $54$.

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Is there any fast way to get the remainders of the division of large numbers?

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$$\frac{77^{200}}{54}$$

Welcome to Mathematics Stack Exchange. By Euler's theorem, $77^{200}=(77^{18})^{11}77^2\equiv23^2\bmod54$ — J. W. Tanner, Dec 18 '20 at 18:07

score 1 · Answer 1 · answered Dec 18 '20 at 10:43

1

As $77\equiv23\pmod{54}, 77^n\equiv23^n$

As Carmichael Function $\lambda(54)=18$

and as $(77,54)=1$

and as $200\equiv2\pmod{18}$

$$23^{200}\equiv23^2{\pmod{54}}\equiv-11\equiv-11+54$$

answered Dec 18 '20 at 10:43

lab bhattacharjee

Please don't answer dupes of FAQs (unless there is something novel that can be said, which is highly unlikely for problems like this). – Bill Dubuque Dec 18 '20 at 15:33
totient function $\phi(54)=18$ too – J. W. Tanner Dec 18 '20 at 18:10

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