What are the last two digits of 7^2008 ?
Last 2 Digits
2 Re: Last 2 Digits on Fri Apr 09, 2010 3:04 am
It's a cycle:07,49,43,01,07 ...
since 2008%4 = 0. So it must be 01
since 2008%4 = 0. So it must be 01
3 @indranil on Fri Apr 09, 2010 3:24 am
01 is the correct ans...
Below i have given a try to come up with the same using binomial expansion...
7 ^ (2008) = 7 ^(4*251) = (7^4)^(251) = (2401)^(251) ....
2401 ^ 251 = (2400 + 1)^251
= Summation [ 251(C)X * (2400^X) * (1^(251  X)) ]
(X = 0 to 251)
As it is obvious from the above term that when X >= 1 , each term will be a multiple of 100 as it will contain 2400 as a factor.. the only term which won't have 2400 as a factor is the first term when X = 0 and the value of term will be 1...
Therefore, 7 ^ (2008) = K * 100 + 1.... where K is a natural no...
Below i have given a try to come up with the same using binomial expansion...
7 ^ (2008) = 7 ^(4*251) = (7^4)^(251) = (2401)^(251) ....
2401 ^ 251 = (2400 + 1)^251
= Summation [ 251(C)X * (2400^X) * (1^(251  X)) ]
(X = 0 to 251)
As it is obvious from the above term that when X >= 1 , each term will be a multiple of 100 as it will contain 2400 as a factor.. the only term which won't have 2400 as a factor is the first term when X = 0 and the value of term will be 1...
Therefore, 7 ^ (2008) = K * 100 + 1.... where K is a natural no...
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