HMMT 二月 2008 · 冲刺赛 · 第 22 题

英文解析

22. [ 10 ] For a positive integer n , let θ ( n ) denote the number of integers 0 ≤ x < 2010 such that x − n is
    2009
    ∑
    divisible by 2010. Determine the remainder when n · θ ( n ) is divided by 2010.
    n =0
    ∑
    2009
    Answer: 335 Let us consider the sum n · θ ( n ) (mod 2010) in a another way. Consider then =0
    2 2 2 2
    sum 0 + 1 + 2 + · · · + 2007 (mod 2010). For each 0 ≤ n < 2010, in the latter sum, the term n
    ∑
    2009
    appears θ ( n ) times, so the sum is congruent to n · θ ( n ). In other words,
    n =0
    2009 2009
    ∑ ∑
    (2009)(2009 + 1)(2 · 2009 + 1) 2010
    n · θ ( n ) = n = ≡ ( − 1) · · ( − 1) = 335 (mod 2010) .2
    6 6
    n =0 n =0