<s>[INST] To find last two digits of $2^{100}$ To find last two digits  of $2^{100}$
I have just learned about modular arithmetic and I wanted to solve this problem. I only know about equivalence classes and about  $a=b \pmod n$. I have also learned about multiplication and addition of classes.  Can someone explain to me step by step on how to apply modular arithmetic to this problem?
Thanks
 [/INST] Method 1(using fast power algorithm):
$2^0 \equiv 1 \pmod {100}$
$2^1 \equiv 2 \pmod {100}$
$2^2 \equiv 4 \pmod {100}$
$2^4 \equiv 16 \pmod {100}$
$2^8 \equiv 56 \pmod {100}$
$2^{16} \equiv 36 \pmod {100}$
$2^{32} \equiv 96 \pmod {100}$
$2^{64} \equiv 16 \pmod {100}$
Since $100 = 4 + 32 + 64$
$2^{100}  \equiv 2^{4} 2^{32} 2^{64}\equiv 16 \times 96 \times 16 \equiv 76 \pmod {100}$
Method 2(using minimal cycle to improve):
Since $2^{22} \equiv 4 \pmod {100}$, $2^{100}  \equiv 2^{22 \times 4 + 12}  \equiv 4^4 \times 2^{12}  \equiv 2^{20} \equiv 16 \times 96 \equiv 76 \pmod{100} $
Last but not least: Since $2^{20} \equiv 76 \pmod {100}$, $76 \times 76 \equiv 76 \pmod{100}$, $2^{100} = 2^{20 \times 5} \equiv 76^5 \equiv 76 \pmod{100} $ by induction. 
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