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atalakumar (0)

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what is the last three digits of 3 to the power 100. how to solve it
    
krishna.gopal (2154)

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3100=950=(10-1)50
150-50C110+50C2102+terms containing higher powers of ten
So last three digits of 3100 will be same as last three digits of 150-50C110+50C2102
150-50C110+50C2102 = 122001
So last three digits of 3100=001

Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi
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