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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