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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.