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.