HMMT 二月 2010 · 代数 · 第 3 题

英文解析

3. [ 4 ] Let S = 0 and let S equal a + 2 a + . . . + ka for k ≥ 1. Define a to be 1 if S < i and -1 if
   0 k 1 2 k i i − 1
   S ≥ i . What is the largest k ≤ 2010 such that S = 0?
   i − 1 k
   Answer: 1092 Suppose that S = 0 for some N ≥ 0. Then a = 1 because N + 1 ≥ S . The
   N N +1 Nfollowing table lists the values of a and S for a few k ≥ N :
   k kk a Sk k
   N 0
   N + 1 1 N + 1
   N + 2 1 2 N + 3
   N + 3 − 1 N
   N + 4 1 2 N + 4
   N + 5 − 1 N − 1
   N + 6 1 2 N + 5
   N + 7 − 1 N − 2
   We see inductively that, for every i ≥ 1,
   S = 2 N + 2 + i
   N +2 iand
   S = N + 1 − i
   N +1+2 ithus S = 0 is the next k for which S = 0. The values of k for which S = 0 satisfy the recurrence
   3 N +3 k krelation p = 3 p +3, and we compute that the first terms of the sequence are 0 , 3 , 12 , 39 , 120 , 363 , 1092;
   n +1 nhence 1092 is our answer.
   Algebra Subject Test