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HMMT 二月 2003 · GEN2 赛 · 第 3 题

HMMT February 2003 — GEN2 Round — Problem 3

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. How many positive rational numbers less than π have denominator at most 7 whenwritten in lowest terms? (Integers have denominator 1.)
解析

英文解析

  1. How many positive rational numbers less than π have denominator at most 7 whenwritten in lowest terms? (Integers have denominator 1.)
    Solution: 54
    We can simply list them. The table shows that there are 3+3+6+6+12+6+18 = 54.
    Denominator Values
    1 2 3
    1 , ,
    1 1 1
    1 3 5
    2 , ,
    2 2 2
    1 2 4 5 7 8
    3 , , , , ,
    3 3 3 3 3 3
    1 3 5 7 9 11
    4 , , , , ,
    4 4 4 4 4 4
    1 2 3 4 6 7 8 9 11 12 13 14
    5 , , , , , , , , , , ,
    5 5 5 5 5 5 5 5 5 5 5 5
    1 5 7 11 13 17
    6 , , , , ,
    6 6 6 6 6 6
    1 2 6 8 9 13 15 16 20
    7 , , . . . , , , , . . . , , , , . . . ,
    7 7 7 7 7 7 7 7 7