2^2008

Note.......2^100  ends in 76

And  [ ....... 76]^n   where n  is a positive integer always ends in 76 

So

2^2000  =   [2^100]^20  = [ ........76]^20  will also end in 76

And 2^8  ends  in 56

So........2^2008   = 2^2000 *2^8 =    [  ...........76] * [ 256]    =  ends in 56

What is the last two digits of the number 2^2008

$\begin{array}{rcll} 2^{2008} \pmod {100} = \ ? \end{array}$

$\boxed{~ \begin{array}{rcll} 2^{32} & \equiv & 96 \pmod {100}\\ & \equiv & 96 -100 \pmod {100}\\ & \equiv & -4 \pmod {100}\\ \end{array} ~}$

$\begin{array}{rcll} 2008 &=& 32\cdot 62 + 24\\\\ & & 2^{2008} \pmod {100} \\ & \equiv & 2^{32\cdot 62 + 24} \pmod {100} \\ & \equiv & (2^{32})^{62}\cdot 2^{24} \pmod {100} \\ & \equiv & (-4)^{62}\cdot 2^{24} \pmod {100} \qquad & |\qquad (-1)^{62} = 1^{62}\\ & \equiv & 4^{62}\cdot 2^{24} \pmod {100} \qquad & |\qquad 2^{24} \pmod {100} = 16\\ & \equiv & 4^{62}\cdot 16 \pmod {100} \\ & \equiv & 4^{62}\cdot 4^2 \pmod {100} \\ & \equiv & 4^{64} \pmod {100} \\ & \equiv & 2^{128} \pmod {100} \qquad & |\qquad 128 =32\cdot 4\\ & \equiv & 2^{32\cdot 4} \pmod {100} \\ & \equiv & (2^{32})^{4} \pmod {100} \\ & \equiv & (-4)^{4} \pmod {100} \qquad & |\qquad (-1)^{4} = 1^{4}\\ & \equiv & 4^4 \pmod {100}\\ & \equiv & 256 \pmod {100}\\ & \equiv & 56 \pmod {100} \end{array}$

the last two digits of the number 2^2008 is 56