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How do you solve this?

$$2^{5^{2017}} \bmod 11$$

I tried my own method, and I got that:

$$2^{5^{2017}} \equiv 10\pmod{11}$$

But I'm not sure it's the correct answer.

Thanks!

Bill Dubuque
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Nello
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    Please show your method and explain why you are unsure about it. – Bill Dubuque Feb 19 '21 at 23:53
  • First, I applied FLT to 5^(2017) mod 11, and got that 78125 ≡ 1(mod 11). Second, I applied FLT to 2^(78125) mod 11, and finally got that the whole thing 2^(5^2017)) ≡ 1 (mod 11) – Nello Feb 20 '21 at 00:19
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    Use the linked mod order reduction method to reduce the expt $\bmod 10,,$ using $$\large 5^{\large 1+n}\bmod 10 = 5(5^{\large n}\bmod 2) = 5(1^{\large n}\bmod 2) = 5(1)\ \ {\rm for},\ n\ge 0$$ by the mod distributive law $\ \ $ – Bill Dubuque Feb 20 '21 at 00:49

1 Answers1

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Note that since $11$ is a prime, $2^{10}\overset{11}{\equiv}1$ Hence because $5^{2017}\overset{10}{\equiv}5$, we have $2^{5^{2017}}\overset{11}{\equiv}2^5\overset{11}{\equiv}10$ and you are right!

Aryan
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