Find the last two digits of $3^{45}$

Essentially, we need to find $\displaystyle 3^{45}\pmod{100}$

Method $\#1:$

Now, $$3^{45}=3\cdot3^{44}=3(3^2)^{22}=3(10-1)^{22}=3(1-10)^{22}$$

Now, $$(1-10)^{22}\equiv1-10\cdot22\pmod{100}\equiv-19\equiv81$$

Method $\#2:$

As $(3,10)=1$

using Euler's Totient Theorem, $\displaystyle \phi(100)=40\implies 3^{40}\equiv1\pmod{100}$

or using Carmichael Function, $\displaystyle \lambda(100)=20\implies 3^{20}\equiv1\pmod{100}$

In either case, $\displaystyle3^{45}\equiv3^5\pmod{100}$