What will be the remainder? I'm stuck with this problem I'm trying to solve from about an hour. Here's the question.
What is the remainder when (3^202)+137 is divided by 101?
There are 4 options -> 36, 45, 56, 11
I want to know the answer of the question with proper and possibly easiest method to solve the problem.
Thanks in advance, waiting for reply. :)

HINT $\  101\ $ is prime so a little Fermat $\rm\ \Rightarrow\ \ 3^{101}\ \equiv\ 3\ \ \Rightarrow\ \ 3^{202}\ \equiv\ \ldots\ (mod\ 101)$
Since your comment reveals you are not familiar with modular arithmetic, here is an alternative.  
By Fermat's little theorem $101$ divides $\: 3^{101}-3\: $ so it also divides $\rm\ (3^{101}-3)\ (3^{101}+3)\ =\ 3^{202}-9$
