HMMT 二月 2003 · 代数 · 第 5 题
HMMT February 2003 — Algebra — Problem 5
题目详情
英文原题
- 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
英文解析
- 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
Solution: 20021
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 havejust 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 thegrounds that the use of the words “several” and “their” in the problem statementimplies there must be at least 2 numbers. Then the answer is 2001, and this maximumis achieved by any two numbers whose sum is 2003.me way.) 3