(2^2013)when divided by 17, the remainder is?

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Ashok Kumar Sarvepalli

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First find the number of cyles for each number. That is 2^1 is 2 , 2^2 is 4 2^3 is 8 2^4 is 16 (6 is the unit digit) 2^5 is 32 ( 2 is the unit digit). So the cycle for the last digitis is ( 2, 4, 8, 6) so for 2^2013 can be written as 2^2012 * 2 = 2^4^504 = 16^504 so it is (16^504 * 2) /17 so (16*16*16*....504...
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First find the number of cyles for each number. That is 2^1 is 2 , 2^2 is 4 2^3 is 8 2^4 is 16 (6 is the unit digit) 2^5 is 32 ( 2 is the unit digit). So the cycle for the last digitis is ( 2, 4, 8, 6) so for 2^2013 can be written as 2^2012 * 2 = 2^4^504 = 16^504 so it is (16^504 * 2) /17 so (16*16*16*....504 times) * 2/17 = 16 divided by 17 is -1 so -1 * -1 * -1...504 times gives 1 so 1 * 2 = 2 read less
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Remainder is 15.
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The required remainder will be 15.
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wrongly explain...when a term is like this ( 16*16*16......504 time)/17 .how can all sixteen's can be divided by only one seventeen..............though the answer is write but it could be better expressed by exponential theorem.
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