Maths Olympiad Prep

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Number theory Difficulty 5.8 AIME, harder Prove it United States

Problem:

It is midnight on April 29th, and Abigail is listening to a song by her favorite artist while staring at her clock, which has an hour, minute, and second hand. These hands move continuously. Between two consecutive midnights, compute the number of times the hour, minute, and second hands form two equal angles and no two hands overlap.

Solution

Solution:

Let t[0,2]t \in [0,2] represent the position of the hour hand, i.e., how many full revolutions it has made. Then, the position of the minute hand is 12t12 t (it makes 12 full revolutions per 1 revolution of the hour hand), and the position of the second hand is 720t720 t (it makes 60 full revolutions per 1 revolution of the minute hand). Then, in order for equal angles to be formed, we need (ab)(bc)=a2b+c0(mod1)(a-b)-(b-c) = a - 2b + c \equiv 0 \pmod{1}, where a,b,ca, b, c is a permutation of t,12t,720tt, 12 t, 720 t. (Here, bb would correspond to the hand that's the angle bisector.) Checking all three possibilities,

12t2(t)+720t697t0(mod1)t2(12t)+720t730t0(mod1)t2(720t)+12t1427t0(mod1) \begin{aligned} 12 t - 2(t) + 720 t &\equiv 697 t \equiv 0 \pmod{1} \\ t - 2(12 t) + 720 t &\equiv 730 t \equiv 0 \pmod{1} \\ t - 2(720 t) + 12 t &\equiv -1427 t \equiv 0 \pmod{1} \end{aligned}

Then we require tt to be a multiple of 1697\frac{1}{697}, 1730\frac{1}{730}, or 11427\frac{1}{1427}. Since 697, 730, and 1427 are pairwise relatively prime, the possible values of tt are

1697,2697,,696697,698697,,26971697,1730,2730,,729730,731730,,27301730,11427,21427,,14261427,14281427,,2142711427 \begin{array}{r} \frac{1}{697}, \frac{2}{697}, \ldots, \frac{696}{697}, \frac{698}{697}, \ldots, \frac{2 \cdot 697 - 1}{697}, \\ \frac{1}{730}, \frac{2}{730}, \ldots, \frac{729}{730}, \frac{731}{730}, \ldots, \frac{2 \cdot 730 - 1}{730}, \\ \frac{1}{1427}, \frac{2}{1427}, \ldots, \frac{1426}{1427}, \frac{1428}{1427}, \ldots, \frac{2 \cdot 1427 - 1}{1427} \end{array}

since t[0,2]t \in [0,2]. This gives a count of 2((6971)+(7301)+(14271))=57022((697-1)+(730-1)+(1427-1)) = 5702.

Note that in the above count we don't count t=0,1,2t = 0, 1, 2 since then all three hands would overlap. If two hands overlap, then one of 11t,708t,719t0(mod1)11 t, 708 t, 719 t \equiv 0 \pmod{1}, and the only way one of these can happen and tt being a multiple of 1697\frac{1}{697}, 1730\frac{1}{730}, or 11427\frac{1}{1427} is if t=12t = \frac{1}{2} and t=32t = \frac{3}{2} (which correspond to 6:00 AM and PM). This is because the only pair of numbers that are not relatively prime among 11, 708, 719, 697, 730, 1427 is 708 and 730. The only common divisor of these two numbers is 2, hence t=12,32t = \frac{1}{2}, \frac{3}{2}. Thus the final answer is 57022=57005702 - 2 = 5700.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.