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How often, and when, do the minute & hour hands overlap?
Well, the first will certainly be at the starting time of midnight (00:00:00).

It's tempting to assume that the next will be at 'five-past-one', then 'ten-past-two', 'quarter-past three' etc., making 24 occurrences per day. But not so.

Consider that by the time the minute hand has reached the 5-minute mark after 1 o'clock, the hour hand will have moved fractionally on.
It's the slow advancement of the hour hand throughout the day that makes the above assumption incorrect. So:

  • when they do they next overlap after midnight?
  • when will they overlap for the 6th time?
  • how many occasions will there be during a day?
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Answers
  • The next time they overlap after midnight is at 1hr 5 5/11 mins (01:05:27).
  • The hands overlap for the 6th time at 05:27:16.
  • The total number of occasions the hands overlap during 24hrs is 22.
Times that the hands overlap:

Notes:
  • Remember that times are quoted to the nearest second.
    The maximum potential time error in displayed results is thus ± ½ second, equating to a a maximum potential angular error between the minute & hour hands of 0.046°.
  • For example, puzzle 5 shows that, at the first overlapping position following midnight listed here as 01:05:27, the actual minute/hour hand angle at that time is 0.025°. While close enough to zero for clocks whose third hand only moves at second intervals, the true point of overlap would in fact be at 01:05:27 3/11.
  • Indeed, there are only two occasions when the angle can be precisely 0°, namely 00:00:00 & 12:00:00 (and the only times the second hand also exactly coincides).
  • In these puzzles the equations for times and angles are absolutely precise, but time results are then rounded to the nearest second, and angle results to either the nearest degree or 1000th of a degree depending on context.
  • If the question is instead:  "When, after any chosen time, do the hands first overlap?", see puzzle 8.
  • If the question is instead:  "When, after any chosen time, will the chosen angle between them be 0°?", see puzzle 9.
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Maths
Method 1 (see Basics):
The angular difference D between overlapping hands is obviously zero so, being less than 180°, the minute/hour hand internal angle A = |5.5M - 30H| = 0, from which 5.5M = 30H. Each occurrence will thus be at M = 30H × 2/11, where M is the minutes after each hour mark H.
The initial 12 o'clock midnight position is properly regarded as 0 o'clock (0 hrs, 00:00:00), so:
  • during the first hour (H = 0):  M = (30 × 0) × 2/11 = 0 mins, i.e. the hands overlap at the start.
  • during the second hour (H = 1):  M = (30 × 1) × 2/11 = 60/11, = 5 5/11, or 5mins 27secs†.
    The time is thus 1hr + (5mins 27secs) = 01:05:27.
Such positions will continue to occur at the above recurrent interval 'IR' of 1hr + 60/11mins = 65 5/11 mins, and the Nth will be at (N - 1) × IR mins.
So, for N = 6:  the time will be 5 × 65 5/11 = 327 3/11 mins, = 5hrs 27mins 6secs† (05:27:16).

Note that for H = 11, M = (30 × 11) × 2/11 = 660/11 = 60 mins, a whole hour, bringing the time to 12:00:00. No overlapping position can therefore occur during the 11th hour. The same is true during the second half of the day, so the hands will overlap on a total of only 22 occasions.
This can be quickly found, without need to calculate them all, by dividing the minutes per day by IR: total N = 1440 / 65 5/11, = 22.

NB:  Be careful to use the precise figure of 65 5/11 mins for IR, rather than that derived from times rounded to the nearest second such as 01:05:27 - 00:00:00. This would otherwise imply IR = 3927 secs, such that for N = 22, the time would be 21 × 3927 = 82467 secs = 22hrs 54mins 27secs, = 22:54:27, 6 seconds short of the true time (to the nearest second) of 22:54:33. It would also wrongly imply a possible 23rd occasion at 23:59:54 rather than it actually being at 24:00:00 (00.00.00 next day).

Method 2 (see Basics):
After the first occasion of zero angular difference at the start, subsequent occurrences will be after the cumulative difference C amounts to 360° and multiples thereof, hence:
  • for N = 1:  C = 0.
  • for N = 2:  C = (360 × 1) = 360.
  • for N = 3:  C = (360 × 2) = 720 etc...
  • it follows that C = 360 × (N - 1), and since C = T × 11/120, T = 360(N - 1) × 120/11 secs.
So, for N = 6:  T = (360 × 5) × 120/11 = 216000/11, = 19636 secs† = 05:27:16.

By substituting the number of seconds per day (86400) for T and solving for N, the number of occurrences per day can be predicted:
86400 = 360(N - 1) × 120/11, so N = ((86400 × 11/120) / 360) + 1. However, the time once 86400 secs have passed is 24:00:00 = 00:00:00 the next day. Since this doesn't count, we can remove the '+ 1' term, so: total N = (86400 × 11/120) / 360, = 22.

† = to the nearest second

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