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{-
- Solution to Project Euler problem 120
- Copyright (c) Project Nayuki. All rights reserved.
-
- https://www.nayuki.io/page/project-euler-solutions
- https://github.com/nayuki/Project-Euler-solutions
-}
{-
- For a given a, what is the n that maximizes the remainder, and what is the value of this remainder?
-
- Let's simplify one term, mod a^2:
- (a+1)^n = 1^n + (n choose 1) a 1^(n-1) + (n choose 2) a^2 1^(n-2) + ... (by the binomial theorem)
- = 1 + an + 0. (remaining addends are 0 because they have a to the power of 2 or more, mod a^2)
- Similarly for the other term, mod a^2:
- (a-1)^n = (-1)^n + (n choose 1) a (-1)^(n-1) + ...
- = (-1)^(n-1) (-1 + an + 0)
- = if n is even then (1 - an) else (an - 1).
- Therefore, adding the two terms:
- (a+1)^n + (a-1)^n
- = if n is even then 2 else 2an.
-
- We can always make 2an >= 2 by taking n=1, for example. So we can disregard the "n is even" case.
- Maximizing 2an mod a^2 for n is the same as maximizing 2n mod a for n.
- If a is even, then the maximum achievable value is a - 2 by setting n = a/2 - 1.
- Else a is odd, then the maximum achievable value is a - 1 by setting n = (a - 1) / 2.
-
- In conclusion, if a is even, the maximum remainder is a(a-2);
- otherwise a is odd, the maximum remainder is a(a-1).
-}
main = putStrLn (show ans)
ans = sum [a * (a - (if (mod a 2) == 0 then 2 else 1)) | a <- [3..1000]]