PUMaC 2016 · 代数（A 组） · 第 4 题

PUMaC 2016 — Algebra (Division A) — Problem 4

Suppose that P is a polynomial with integer coefficients such that P (1) = 2, P (2) = 3 and P (3) = 2016. If N is the smallest possible positive value of P (2016), find the remainder when N is divided by 2016. 2

解析

Using the well known result that x − x | P ( x ) − P ( x ) we get that N ≡ 2 (mod 2015), N ≡ 3 1 2 1 2 (mod 2014) and N ≡ 3 (mod 2013). Solving these equations gives N ≡ 3 + 1007 × 2013 × 2014 (mod 2013 × 2014 × 2015). Thus N = 3 + 1007 × 2013 × 2014 and N ≡ 3 + 1007 × ( − 3) × ( − 2) ≡ 2013 (mod 2016). The polynomial P ( x ) = ( x − 1)(( x − 2)(1006) + 1) + 2 satisfies all given conditions, so 2013 is the final answer. Problem written by Mel Shu. 12 k k − 1 k +1 k − 1