My question is to find the last two digits of $3^{400}$ and I found that the last digit is $1$ but I'm not sure about the other one. Any help would be appreciated, thank you for your time. Edit: I've found that $3^{\phi(100)} = 1 \pmod{100} ; \phi(100) = 40$ and $3^{400} = 1 \pmod{100}$ So the last digit of $3^{400}$ is $1$
I am trying to calculate the last two digits of $3^{400}$, but working $\bmod {100}$ is proving intractable. I did the last digit case pretty handily, and now I'm stuck at last two.
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