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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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