1.

Find the last three digits in `11^(50)`

Answer» Expansion of `(10 +1)^(50)=""^(50)C_010^(50)+""^(50)C_(1)10^(49)+…..+""^(50)C_(48)10^2+""^(50)C_(49)10+""^(50)C_(50)`
`= ubrace(""^(50)C_(0)10^(50)+""^(50)C_(1)10^(49)+.......+""^(50)C_(47)10^(3))_("1000k")+49xx25xx100+500+1`
`rArr 1000k +123001`
`rArr` Last 3 digits are 001


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