I am trying to calculate 10130mod48 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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