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by anshumishra » Thu Dec 30, 2010 3:48 pm
Night reader wrote:What is the last digit of (1,103 - 816)^39

A. 3
B. 4
C. 5
D. 7
E. 9
A

Last digit of 287^39 = Last digit of 7^39
Power of 7 has a cycle of 4 (1,9,3,4,....repeat).

So, L[7^(4*9+3)] = L(7^3) = 3.
Thanks
Anshu

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by Night reader » Thu Dec 30, 2010 3:55 pm
anshumishra wrote:
Night reader wrote:What is the last digit of (1,103 - 816)^39

A. 3
B. 4
C. 5
D. 7
E. 9
A

Last digit of 287^39 = Last digit of 7^39
Power of 7 has a cycle of 4 (1,9,3,4,....repeat).

So, L[7^(4*9+3)] = L(7^3) = 3.
Absolutely fine answer, thanks Anshu!
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