What are the last two digits in the ordinary decimal representation of $3^{400}$ 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 $1$

The multiplicative group of integers mod 100 has order of $\phi (100) = 100 \cdot \frac 12 \cdot \frac 45 = 40$
$a^{40} \equiv 1 \pmod {100}\\
(a^{40})^{10} \equiv 1 \pmod {100}$
The last two digits are $01$
