HMMT 二月 2015 · 冲刺赛 · 第 10 题

HMMT February 2015 — Guts Round — Problem 10

[ 6 ] Let b ( x ) = x + x + 1. The polynomial x + x + · · · + x + 1 has a unique “base b ( x )” representation N ∑ 2015 2014 k x + x + · · · + x + 1 = a ( x ) b ( x ) , k k =0 where • N is a nonnegative integer; • each “digit” a ( x ) (for 0 ≤ k ≤ N ) is either the zero polynomial (i.e. a ( x ) = 0) or a nonzero k k polynomial of degree less than deg b = 2; and • the “leading digit a ( x )” is nonzero (i.e. not the zero polynomial). N Find a (0) (the “leading digit evaluated at 0”). N

解析

[ 6 ] Let b ( x ) = x + x + 1. The polynomial x + x + · · · + x + 1 has a unique “base b ( x )” representation N ∑ 2015 2014 k x + x + · · · + x + 1 = a ( x ) b ( x ) , k k =0 where • N is a nonnegative integer; • each “digit” a ( x ) (for 0 ≤ k ≤ N ) is either the zero polynomial (i.e. a ( x ) = 0) or a nonzero k k polynomial of degree less than deg b = 2; and • the “leading digit a ( x )” is nonzero (i.e. not the zero polynomial). N Find a (0) (the “leading digit evaluated at 0”). N Answer: − 1006 Comparing degrees easily gives N = 1007. By ignoring terms of degree at most 2013, we see 2 1007 2015 2014 2013 a ( x )( x + x + 1) ∈ x + x + O ( x ) . N Write a ( x ) = ux + v , so N 2 1007 2014 2013 2012 a ( x )( x + x + 1) ∈ ( ux + v )( x + 1007 x + O ( x )) N 2015 2014 2013 ⊆ ux + ( v + 1007 u ) x + O ( x ) . Finally, matching terms gives u = 1 and v + 1007 u = 1, so v = 1 − 1007 = − 1006. Remark. This problem illustrates the analogy between integers and polynomials, with the nonconstant 2 (degree ≥ 1) polynomial b ( x ) = x + x + 1 taking the role of a positive integer base b > 1.