what are the last two digits of 2 raised to the power of 64?please give me the solution too.thanks.

One can observe

Last 2 digits of

24^even is 76

24^odd is 24

So

2^64=2^60*2^4

=(1024)^6*2^4

So last 2 digits are 76*16=1216

So 16 is the answer

All the best.

AKASH GOYAL

AskiitiansExpert-IITD

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