Lots of 2013s
You have 2013 2013's, and you arrange them such that you create the massive number 2 0 1 3 2 0 1 3 2 0 1 3 2 0 1 3 . . . 2 0 1 3 . You call this number ’Mr. Huge’ . Find ’Mr. Huge’ ( m o d 8 ) .
The answer is 5.
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2 solutions
= = = = = = 2 0 1 3 2 0 1 3 2 0 1 3 ⋯ 2 0 1 3 ( 2 0 1 3 m o d 8 ) 2 0 1 3 2 0 1 3 ⋯ 2 0 1 3 5 2 0 1 3 2 0 1 3 ⋯ 2 0 1 3 5 ( 2 0 1 3 m o d 4 ) 2 0 1 3 ⋯ 2 0 1 3 5 1 2 0 1 3 ⋯ 2 0 1 3 5 1 5 m o d 8 m o d 8 m o d 8 m o d 8 m o d 8 m o d 8 Euler’s Totient Theorem
2013 is congruent to 5 modulo 8.Also, ϕ ( 8 ) = 4 and 2013 is congruent to 1 modulo 4.So the answer is 5