TITLE: Finding the last two digits $123^{562}$
QUESTION [6 upvotes]: This is my question:
Find the last 2 digits of $123^{562}$
I don't even understand where to begin on this problem. I know I'm supposed to use Euler's theorem but I have no idea how or why. Any help? Thanks

REPLY [2 votes]: I think it will be 29 . You can do it in this way...... $123^{562}=23^{562}=(562*3^{561}*20+3^{562})=29$ mod 100 because $3^{560}=81^{140}=(140*80*1+1)$ mod 100 so $3^{561}=03$ and $3^ {562}=09$ mod 100 . In all the steps i have used binomial theorem.