Let P ( x ) be a monic polynomial of degree 2 0 1 5 , satisfying, for x ∈ Z + , x ≤ 2 0 1 5 , P ( x ) = k = 1 ∑ x k 3 + 1 Q ( k ) 3 + 1 where Q ( x ) = ⎩ ⎪ ⎨ ⎪ ⎧ − i = 1 ∑ ∞ i ∣ x ∣ 1 if the sum converges − ∣ x ∣ 1 if the sum diverges Now, let a n be the coefficient of the x n term in P ( x ) . Finally, let S = i = 1 ∑ 2 0 1 5 a i − 1 Find the positive remainder when ⌊ S ⌋ 2 0 1 5 − ⌈ S ⌉ is divided by 1 0 0 0 .
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This is a better trolling attempt than my problem , although definitely much less obvious. Well done.
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I think you're problem is better; I still don't see how to do it at a glance. I will look into it deeper when I have the time.
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Note that i = 0 ∑ 2 0 1 5 a i = P ( 1 ) and a 2 0 1 5 = 1 so S = P ( 1 ) − 1
But P ( 1 ) = 2 Q ( 1 ) 3 + 1 = 2 − 1 + 1 = 0 so S = − 1 .
Finally, ⌊ S ⌋ 2 0 1 5 − ⌈ S ⌉ = ( − 1 ) 2 0 1 6 = 1 and taking mod 1 0 0 0 we get the answer as 1 .