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Duke Math Meet 2008 Problem 8 Solution (Individual Round)

Find the last two digits of \(\sum_{k=1}^{2008} k {{2008}\choose{k}} \)

Discussion:

\((1+x)^n = \sum_{k=0}^{n} {{n}\choose{k}}x^k \)

We differentiate both sides to have \(n(1+x)^{n-1} = \sum_{k=1}^{n} k {{n}\choose{k}}x^{k-1} \)

Put n=2008 and x=1in the above equation to have

\(2008(1+1)^{2007} = \sum_{k=1}^{2008} k {{2008}\choose{k}}1^{k-1} \)

Thus \(\sum_{k=1}^{2008} k {{2008}\choose{k}} = 2^{2007} \times 2008 \equiv 8 \times 2^{2007} \equiv 2^{2010} \) mod 100

Now \(2^{12} = 4096 \equiv -4 \) mod 100 . Now raising both sides to the power of 6, we have \(2^{72} \equiv (-4)^6 \equiv 2^{12} \equiv -4 \)mod 100 . Again raising both sides to the power 6 we have \(2^{432} \equiv -4 \Rightarrow 2^{1728} \equiv 2^8 \equiv 56 \) mod 100.

Earlier we had \(2^{72} \equiv -4 \Rightarrow 2^{216} \equiv -64 \equiv 36 \) mod 100

Hence \(2^{1728} \times 2^{216} \equiv 2^{1944} \equiv 56 \times 36 \equiv 16 \) mod 100

Finally we know \(2^{12} \equiv -4 \Rightarrow 2^{60} \equiv -1024 \equiv 76 \) mod 100 . Hence \(2^{1944} \times 2^{60} \equiv 2^{2004} \equiv 16 \times 76 \equiv 1216 \equiv 16 \) mod 100

Thus \(2^{2004} \times 2^{6} \equiv 2^{2010} \equiv 16 \times 64 \equiv 1024 \equiv 24 \) mod 100.

The two digits are 24.

October 15, 2014

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