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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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