Last Digits

A short outline of the solution...

Note

To find last two digits , we always take $\mod 100$

Using the Euler's Theorem , which is , $a^{\phi (n)} \equiv 1 \mod n$ . Applying it in the problem ...we have , $\phi (100) = 40$ . So we arrive at a congruence which is $2^{12} \mod 100$ . If you are wondering how that 12 came to the power of base 2 ,

then note that 12 is the remainder when we divide 2012 by 40.

So the final congruence is , $4096 \mod 100$ and hence the last two digits are 96

Thanks to Euler , who made it all easy...

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Last 2 digits for powers of 2 cycles for every 20 and so we can find the last 2 digit directly for $2^{12}$ = 1024 * 4 =96