I tried writing out the first few powers of 3 to see if I could find any pattern in the last two digits:  $3^0=01$  $3^1=03$  $3^2=09$  $3^3=27$  $3^4=81$  $3^5=43$, etc.  However, I came up empty-handed, and I am unsure how to approach this.   Any ideas? Many thanks.
I was wondering if there is a simpler way to find the last to digits of a power such as $3^{45}$. I reduced it modulo 100 to get the answer, which is 43. But I was curious if there was a simpler, or more eloquent way to get this result.  Here's my method:  Using the fact that $3^4=81\equiv -19 \pmod{100}$ and $19^5=2,476,099\equiv -1 \pmod{100}$  $$ 3^{45}\equiv 3(3^4)^{11}\equiv 3(-19)^{11}\equiv 3(-19)(19^5)^2\equiv 3(-19)(-1)^2 \equiv -57 \equiv 43 \pmod {100} $$  I just felt like this was sort of a messy way to find out that the last two digits are 43.
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