Let’s try to find the solution to Duke Math Meet 2008 Problem 8. This question is from the individual round of that meet.
Problem: Duke Math Meet 2008 Problem 8
Find the last two digits of
We differentiate both sides to have
Put n=2008 and x=1in the above equation to have
Thus mod 100
Now mod 100 . Now raising both sides to the power of 6, we have mod 100 . Again raising both sides to the power 6 we have mod 100.
Earlier we had mod 100
Hence mod 100
Finally we know mod 100 . Hence mod 100
Thus mod 100.
The two digits are 24.