HMMT 二月 2015 · 冲刺赛 · 第 10 题

英文解析

10. [ 6 ] Let b ( x ) = x + x + 1. The polynomial x + x + · · · + x + 1 has a unique “base b ( x )”
    representation
    ∑N
    2015 2014 kx + x + · · · + x + 1 = a ( x ) b ( x ) ,
    k =0 kwhere
    • 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 nonzerok kpolynomial of degree less than deg b = 2; and
    • the “leading digit a ( x )” is nonzero (i.e. not the zero polynomial).
    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 n
    2013, we see
    2 1007 2015 2014 2013
    a ( x )( x + x + 1) ∈ x + x + O ( x ) .
    Write a ( x ) = ux + v , so n
    2 1007 2014 2013 2012Na ( x )( x + x + 1) ∈ ( ux + v )( x + 1007 x + O ( x ))
    2015 2014 2013N
    ⊆ 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
    (degree ≥ 1) polynomial b ( x ) = x + x + 1 taking the role of a positive integer base b > 1.2