1@joydoot : nice solution :)
Here is my solution using elementary and pure number theory
52 ≡ 25 mod 13
≡ -1 mod 13
(52)49 ≡ (-1)49 ≡ -1 mod 13...............1
and 5 = 5 mod 13................2
multiplying 1 and 2
599 ≡ -5 ≡ 8 mod 13
:)
3 Answers
115.(5)98
=5.(25)49
=5.(26-1)49.
expand binomially,
5.[ 49C0(26)49 - 49C1(26)48.............. +49C48(26) - 49C49]
26 is a factor of 13.
therefore the remainder is..
13-5 = 8. :)
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