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HMMT 二月 2006 · GEN2 赛 · 第 9 题

HMMT February 2006 — GEN2 Round — Problem 9

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, jumpingon the minute). He also has a clock that runs really slow—at half the correct rate, to be exact. Atnoon 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 angleson the two clocks equal?
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 rangesin [ − 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.