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HMMT 二月 2006 · 代数 · 第 5 题

HMMT February 2006 — Algebra — Problem 5

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Tim has a working analog 12-hour clock with two hands that run continuously (instead of,
    say, jumping on the minute). He also has a clock that runs really slow—at half the correctrate, to be exact. At noon one day, both clocks happen to show the exact time. At any given
    ° °
    instant, the hands on each clock form an angle between 0 and 180 inclusive. At how manytimes during that day are the angles on the two clocks equal?

    √ √
    3 2 4 2
解析

英文解析

  1. Tim has a working analog 12-hour clock with two hands that run continuously (insteadof, say, jumping on the minute). He also has a clock that runs really slow—at halfthe correct rate, to be exact. At noon one day, both clocks happen to show the exact
    °
    time. At any given instant, the hands on each clock form an angle between 0 and
    °
    180 inclusive. At how many times during that day are the angles on the two clocksequal?
    Answer: 33
    Solution: A tricky thing about this problem may be that the angles on the twoclocks might be reversed and would still count as being the same (for example, both
    °
    angles could be 90 , but the hour hand may be ahead of the minute hand on one clockand behind on the other).
    °
    Let x , − 12 ≤ x < 12, denote the number of hours since noon. If we take 0 to meanupwards to the “XII” and count angles clockwise, then the hour and minute hands of
    ° ° °
    the correct clock are at 30 x and 360 x , and those of the slow clock are at 15 x and
    ° ° °
    180 x . The two angles are thus 330 x and 165 x , of course after removing multiples of
    °
    360 and possibly flipping sign; we are looking for solutions to
    ° ° ° ° ° °
    330 x ≡ 165 x (mod 360 ) or 330 x ≡ − 165 x (mod 360 ) .
    In other words,
    360 | 165 x or 360 | 495 x.
    Or, better yet,
    165 11 495 11
    x = x and/or x = x
    360 24 360 8
    must be an integer. Now x is any real number in the range [ − 12 , 12), so 11 x/ 8 ranges 2
    in [ − 16 . 5 , 16 . 5), an interval that contains 33 integers. For any value of x such that
    11 x/ 24 is an integer, of course 11 x/ 8 = 3 × (11 x/ 24) is also an integer, so the answeris just 33.

    √ √
    3 2