HMMT 十一月 2021 · 团队赛 · 第 3 题

HMMT November 2021 — Team Round — Problem 3

[30] Suppose m and n are positive integers for which • the sum of the first m multiples of n is 120, and 3 3 • the sum of the first m multiples of n is 4032000. 2 2 Determine the sum of the first m multiples of n .

解析

[30] Suppose m and n are positive integers for which • the sum of the first m multiples of n is 120, and 3 3 • the sum of the first m multiples of n is 4032000. 2 2 Determine the sum of the first m multiples of n . Proposed by: Sean Li Answer: 20800 Solution: For any positive integers a and b , the sum of the first a multiples of b is b + 2 b + · · · + ab = a ( a +1) b 3 3 3 b (1+2+ · · · + a ) = . Thus, the conditions imply m ( m +1) n = 240 and m ( m +1) n = 8064000, 2 whence 3 3 3 ( m + 1) ( m ( m + 1) n ) 240 12 = = = . 3 3 3 3 m + 1 m ( m + 1) n 8064000 7 2 2 2 Thus, we have 7( m + 1) = 12( m − m + 1) or 5 m − 26 m + 5 = 0, so m = 5 and therefore n = 8. The 2 2 m ( m +1) 2 answer is n = 20800. 2