The remainder when is divided by is
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By Fermat's little theorem, a p ≡ a ( m o d p ) , for any a ∈ N and p be any prime number
Now, putting a = 3 , 5 and p = 1 3 we get the followings,
3 1 3 ≡ 3 ( m o d 1 3 ) and 5 1 3 ≡ 5 ( m o d 1 3 )
⟹ 3 1 2 ≡ 1 ( m o d 1 3 ) and 5 1 2 ≡ 1 ( m o d 1 3 )
and adding these two we get finally
3 1 2 + 5 1 2 ≡ 2 ( m o d 1 3 )
i.e. the remainder is 2 when 3 1 2 + 5 1 2 is divisible by 1 3