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Hands close together

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On an analog clock the hour hand turns $0.5$ degrees per minute and the minute hand $6$ degrees per minute, both starting straight up at $12 : 00$ . Consider the $720$ whole-minute times from $12 : 00$ up to but not including the next $12 : 00$ . At how many of them is the smaller angle between the hands at most $30$ degrees (equal to $30$ counts)?

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Hint

The relative angle grows by $5.5$ degrees per minute; count the minutes around each of the $11$ times the hands line up.

Solution

After $m$ minutes the hands are $5.5 m$ degrees apart modulo $360$ , so the smaller angle is at most $30$ degrees exactly when $11 m$ lies within $60$ of a multiple of $720$ .
Around each alignment, whole minutes $m$ with $∣11 m - 720 k ∣\leq 60$ form a window of $11$ consecutive minutes for $k = 1, 2, ..., 10$ .
The windows for $k = 0$ and $k = 11$ straddle $12 : 00$ : they contain $6$ minutes at the start and $5$ minutes at the end of the period, which together make one more window of $11$ minutes.
There are $11$ windows of $11$ minutes, so the count is $11 \times 11 = 121$ .

Answer: $121$

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