Euler
n = 1 ∑ 1 0 0 8 n 2 0 1 8
Find the remainder when the expression above is divided by 2017?
The answer is 0.
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1 solution
It's worth mentioning that g c d ( n , 2 0 1 7 ) = 1 ∀ n ∈ N : n < 2 0 1 7 and hence Euler's theorem is valid here ∀ n ∈ N : 1 ≤ n ≤ 1 0 0 8 .
Edit: Ah I see that it is enough to just tell that 2 0 1 7 is a prime number.
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Yes, I just deleted it to make the solution shorter.
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It's completely okay sir. Stating that 2017 is a prime is a better shorthand than that (read my edit on my comment).
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@Tapas Mazumdar – I have added the g cd ( n , 2 0 1 7 ) = 1 .
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@Chew-Seong Cheong – Thanks, it seems better to understand now.
Relevant wiki: Euler's Theorem
n = 1 ∑ 1 0 0 8 n 2 0 1 8 ≡ n = 1 ∑ 1 0 0 8 n 2 0 1 8 mod ϕ ( 2 0 1 7 ) (mod 2017) ≡ n = 1 ∑ 1 0 0 8 n 2 0 1 8 mod 2 0 1 6 (mod 2017) ≡ n = 1 ∑ 1 0 0 8 n 2 (mod 2017) = 6 1 0 0 8 ( 1 0 0 9 ) ( 2 0 1 7 ) (mod 2017) = 0 (mod 2017) As 2017 is a prime, g cd ( n , 2 0 1 7 ) = 1 , Euler’s theorem applies. Euler’s totient function ϕ ( 2 0 1 7 ) = 2 0 1 6 Note that k = 1 ∑ n k 2 = 6 n ( n + 1 ) ( 2 n + 1 )