I am trying to calculate $10^{130} \bmod 48$ but I need to use Euler's theorem in the process.
I noticed that 48 and 10 are not coprime so I couldn't directly apply Euler's theorem. I tried breaking it down into $5^{130}2^{130} \bmod 48$ and I was sucessfully able to get rid of the 5 using Euler's theorem but now I'm stuck with $2^{130} \bmod 48$. $2^{130}$ is still a large number and unfortunately 2 and 48 are not coprime.
Could someone lead me in the right direction regarding how would I proceed from here?
Answer
Calculate $\mod 48$ using the Chinese Remainder Theorem. Or, informally:
Clearly $2^{130}$ is divisible by $16$ so modulo $48$ this is one of $0, 16, 32$, which one of the three it is depends on what it is modulo $3$.
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