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HMMT 二月 2003 · 冲刺赛 · 第 36 题

HMMT February 2003 — Guts Round — Problem 36

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [12] A teacher must divide 221 apples evenly among 403 students. What is the minimalnumber of pieces into which she must cut the apples? (A whole uncut apple counts asone piece.)
    HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

英文原题

[12] A teacher must divide 221 apples evenly among 403 students. What is the minimal
number of pieces into which she must cut the apples? (A whole uncut apple counts as
one piece.)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

解析

英文解析

  1. A teacher must divide 221 apples evenly among 403 students. What is the minimalnumber of pieces into which she must cut the apples? (A whole uncut apple counts asone piece.)
    Solution: 611
    Consider a bipartite graph, with 221 vertices representing the apples and 403 verticesrepresenting the students; each student is connected to each apple that she gets apiece of. The number of pieces then equals the number of edges in the graph. Each 9
    student gets a total of 221 / 403 = 17 / 31 apple, but each component of the graphrepresents a complete distribution of an integer number of apples to an integer numberof students and therefore uses at least 17 apple vertices and 31 student vertices. Thenwe have at most 221 / 17 = 403 / 31 = 13 components in the graph, so there are atleast 221 + 403 − 13 = 611 edges. On the other hand, if we simply distribute in thestraightforward manner — proceeding through the students, cutting up a new applewhenever necessary but never returning to a previous apple or student — we can createa graph without cycles, and each component does involve 17 apples and 31 students.
    Thus, we get 13 trees, and 611 edges is attainable.