Online Course Discussion Forum
Math Challenge II-A Spring 2022 Q10.10
We didn't get through the last question, so I was wondering if the final answer is 1?
Since 11^2011 = 1(mod 8)
and 11^2011 = 121(mod 125)
The Chinese Remainder Theorem can manipulate them to 121(mod 1000) which would leave 1 in the hundreds place.
Is this correct?