What are the last two digits of $2^{100}$ ??

Question

I am learning the algebra with remainders. For example: find the remainder of $2^{2011}$ divided by $3$ which is $11$. My problems is when i have lower power number then the divisor for example: $3^{56}$ divided $77$ is confusing for me. Someone help pls...

Answer 1

We have $$\begin{align}2^{100}&\equiv(2^{10})^{10}\equiv(1024\times1024)^5\equiv(24\times24)^5\equiv(576\times576)^2\times576\\&\equiv(76\times76)^2\times76\equiv5776^2\times76\equiv76^3\equiv76\pmod{100}\end{align}$$

Answer 2

Without any advanced tool simply note that $$2^{10}=1024$$ then we can consider $4^{10}=2^{20}=(2^{10})^2$ that is equivalent to consider $$24^2=576$$