5^2011 mod 7 ??? what the answer ??
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By Fermat's little theorem (or just by checking powers of 5)
5^6 = 1 (mod 7)
Therefore
5^2011 = 5^(6*335 + 1)
= (5^6)^335 * 5^1
= 1^335 * 5
= 5 (mod 7).
5^6 = 1 (mod 7)
Therefore
5^2011 = 5^(6*335 + 1)
= (5^6)^335 * 5^1
= 1^335 * 5
= 5 (mod 7).
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Fermat's little theorem (FLT)
7 can't divide 5, 7 is prime so 5^(7-1)= 5^6 = 1 mod 7 <-- FLT
So given: 5^6 = 1 mod 7
and 335*6=2010
5^2011 = 5^2010 * 5 = (5^6)^335 * 5
Question is the same as: (5^6)^335 * 5 (mod 7)
but (5^6)^335 mod 7 = 1 as 5^6 = 1 mod 7
thus, (5^6)^335 * 5 (mod 7) = 5 mod 7 (same as saying the remainder is 5 upon division)
7 can't divide 5, 7 is prime so 5^(7-1)= 5^6 = 1 mod 7 <-- FLT
So given: 5^6 = 1 mod 7
and 335*6=2010
5^2011 = 5^2010 * 5 = (5^6)^335 * 5
Question is the same as: (5^6)^335 * 5 (mod 7)
but (5^6)^335 mod 7 = 1 as 5^6 = 1 mod 7
thus, (5^6)^335 * 5 (mod 7) = 5 mod 7 (same as saying the remainder is 5 upon division)