HMMT 十一月 2023 · 冲刺赛 · 第 30 题

HMMT November 2023 — Guts Round — Problem 30

[15] An HMMT party has m MIT students and h Harvard students for some positive integers m and h , For every pair of people at the party, they are either friends or enemies. If every MIT student has 16 MIT friends and 8 Harvard friends, and every Harvard student has 7 MIT enemies and 10 Harvard enemies, compute how many pairs of friends there are at the party. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2023, November 11, 2023 — GUTS ROUND Organization Team Team ID#

解析

[15] An HMMT party has m MIT students and h Harvard students for some positive integers m and h , For every pair of people at the party, they are either friends or enemies. If every MIT student has 16 MIT friends and 8 Harvard friends, and every Harvard student has 7 MIT enemies and 10 Harvard enemies, compute how many pairs of friends there are at the party. Proposed by: Reagan Choi Answer: 342 Solution: We count the number of MIT-Harvard friendships. Each of the m MIT students has 8 Harvard friends, for a total of 8 m friendships. Each of the h Harvard students has m − 7 MIT friends, for a total of h ( m − 7) friendships. So, 8 m = h ( m − 7) = ⇒ mh − 8 m − 7 h = 0 = ⇒ ( m − 7)( h − 8) = 56. Each MIT student has 16 MIT friends, so m ≥ 17. Each Harvard student has 10 Harvard enemies, so h ≥ 11. This means m − 7 ≥ 10 and h − 8 ≥ 3. The only such pair ( m − 7 , h − 8) that multiplies to 56 is (14 , 4) , so there are 21 MIT students and 12 Harvard students. ( h − 1 − 10) h 16 m We can calculate the number of friendships as + 8 m + = 168 + 168 + 6 = 342 . 2 2