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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 **

Discussion:

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.

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