HMMT 十一月 2023 · THM 赛 · 第 10 题
HMMT November 2023 — THM Round — Problem 10
题目详情
- 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.
解析
- 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. Proposed by: Evan Erickson Answer: 5700 Solution: Let t ∈ [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 12 t (it makes 12 full revolutions per 1 revolution of the hour hand), and the position of the second hand is 720 t (it makes 60 full revolutions per 1 revolution of the minute hand). Then, in order for equal angles to be formed, we need ( a − b ) − ( b − c ) = a − 2 b + c ≡ 0 (mod 1), where a, b, c is a permutation of t, 12 t, 720 t . (Here, b would correspond to the hand that’s the angle bisector.) Checking all three possibilities, 12 t − 2( t ) + 720 t ≡ 697 t ≡ 0 (mod 1) , t − 2(12 t ) + 720 t ≡ 730 t ≡ 0 (mod 1) , t − 2(720 t ) + 12 t ≡ − 1427 t ≡ 0 (mod 1) . 1 1 1 Then we require t to be a multiple of , , or . Since 697, 730, and 1427 are pairwise relatively 697 730 1427 prime, the possible values of t are 1 2 696 698 2 · 697 − 1 , , . . . , , , . . . , , 697 697 697 697 697 1 2 729 731 2 · 730 − 1 , , . . . , , , . . . , , 730 730 730 730 730 1 2 1426 1428 2 · 1427 − 1 , , . . . , , , . . . , 1427 1427 1427 1427 1427 since t ∈ [0 , 2]. This gives a count of 2((697 − 1) + (730 − 1) + (1427 − 1)) = 5702 . Note that in the above count we don’t count t = 0 , 1 , 2 since then all three hands would overlap. If two hands overlap, then one of 11 t, 708 t, 719 t ≡ 0 (mod 1), and the only way one of these can happen 1 1 1 1 3 and t being a multiple of , , or is if t = and t = (which correspond to 6:00 AM and 697 730 1427 2 2 PM). This is because the only pair of numbers that are not relatively prime among 11, 708, 719, 697, 1 3 730, 1427 is 708 and 730. The only common divisor of these two numbers is 2, hence t = , . Thus 2 2 the final answer is 5702 − 2 = 5700.