Let r 1 , r 2 , . . . , r 2 0 1 4 be the complex roots, not all necessarily distinct, of the polynomial x 2 0 1 4 + 2 0 1 4 x 2 0 1 3 − 1 . If
S = n = 1 ∑ 2 0 1 4 k = 1 ∑ 2 0 1 4 r k n ,
and 2 0 1 4 2 0 1 5 S = a 2 0 1 4 + b , for some positive integers a and b such that a is as large as possible, find the units digit of a + b + 2 0 1 4 .
This problem is part of the set Yay for 2014! .
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did not see that unit digit had to be written :(
Didn't notice that I have to answer the units digit only.
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Using Newton's Sums where a(2014) = 1, a(2013) = 2014, a(2012) up to a(1) = 0, and a(0) = -1.
From here, S(1) = -2014 S(2) = 2014^2 ... S(n) = (-2014)^n for 1 <= n <= 2013 S(2014) = 2014^2014 + 2014.
Summing up and multiplying the sum by (2015/2014) gives 2014^2014 + 2014... gives a + b + c = 6042 and hence, 2.