HMMT 二月 2003 · 代数 · 第 5 题

HMMT February 2003 — Algebra — Problem 5

Several positive integers are given, not necessarily all different. Their sum is 2003. Suppose that n of the given numbers are equal to 1, n of them are equal to 2, . . . , 1 2 n of them are equal to 2003. Find the largest possible value of 2003 n + 2 n + 3 n + · · · + 2002 n . 2 3 4 2003 3

解析

Several positive integers are given, not necessarily all different. Their sum is 2003. Suppose that n of the given numbers are equal to 1, n of them are equal to 2, . . . , 1 2 n of them are equal to 2003. Find the largest possible value of 2003 n + 2 n + 3 n + · · · + 2002 n . 2 3 4 2003 1 Solution: 2002 The sum of all the numbers is n + 2 n + · · · + 2003 n , while the number of numbers 1 2 2003 is n + n + · · · + n . Hence, the desired quantity equals 1 2 2003 ( n + 2 n + · · · + 2003 n ) − ( n + n + · · · + n ) 1 2 2003 1 2 2003 = (sum of the numbers) − (number of numbers) = 2003 − (number of numbers) , which is maximized when the number of numbers is minimized. Hence, we should have just one number, equal to 2003, and then the specified sum is 2003 − 1 = 2002. Comment: On the day of the contest, a protest was lodged (successfully) on the grounds that the use of the words “several” and “their” in the problem statement implies there must be at least 2 numbers. Then the answer is 2001, and this maximum is achieved by any two numbers whose sum is 2003.me way.) 3