Triangulating for the simplest puzzle that is still at least somewhat interesting to solve..

On the left side wall in this picture, we have two particular clocks:

1: an analog clock with identical hour and minute arms, and

2: a digital clock that shows initial zeroes when appropriate

During the course of a day, which of these clocks agrees more often with its horizontal mirror image?

enter image description here


The times where the analog clock will match are

The same as the times that the analog clock has reflective symmetry across it's "y" axis. This will happen at least once an hour. The proof for this involves the Intermediate Value Theorem, but essentially boils down to the fact that, every hour, the hour hand continuously sweeps one part of the clock, while the minute hand continuously sweeps the entire clock. It follows that, in sweeping the entire clock, it must for an instant pass through the part of the clock that is currently opposite the hour hand.

It actually matches slightly more often than this, because they're already matching at noon/midnight, so it comes out to 13 matches every 12 hours, or 26 total.

The times where the digital clocks will match are

Again, where the clock has reflective symmetry across it's "y" axis (The colon). As @Dorrulf has already enumerated, this happens at 00:00, 02:50, 20:05, and 22:55. I'm not counting the times with ones in them, because they don't show up on the same half of the digit when reflected. This is a measly 4 total

So, it's clear that the reflections match more often for the

Analog Clock

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    $\begingroup$ There is an additional match at 6:00 and 18:00 on analog clock since both handles are exactly on vertical axis. $\endgroup$ – Phil1970 Oct 24 '18 at 1:19
  • $\begingroup$ This is one way to look at it, but i'd still say two digital clocks with a different font can show the same time, thus agreeing with each other. With that interpretation, you can get a similar bound for the number of occurrences, but this time around, it's not a lower bound. $\endgroup$ – Bass Oct 24 '18 at 4:20
  • $\begingroup$ I think there is a slight mistake for the analog clock when you pass from X times avery 12 hours to 2X times every 24 hours: you are counting noon and midnight twice. However, the total seems correct if we add @Phil1970 's remark (although 'on the vertical axis' was not the best way to phrase it). $\endgroup$ – Evargalo Oct 24 '18 at 6:55
  • $\begingroup$ Fun fact: if the arms of the digital clock were not of the same length, we would have a tie between both clocks. $\endgroup$ – Evargalo Oct 24 '18 at 6:58
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    $\begingroup$ The digital clock also matches at 05:20. $\endgroup$ – Rosie F Oct 28 '18 at 7:51

Assuming 'agrees with' is synonymous with 'looks the same'...


As suggested by DqwertyC, the analogue clock will agree 26 times in 24 hours if you include both mid-nights. If you argue that there is only one midnight in a 24 hour day then this can be reduced to 25 times. Assuming the latter these times are approximately: 00:55, 1:50, 2:46, 3:41, 4:37, 5:32, 6:28, 7:22, 8:18, 9:14, 10:09, 11:04, and 12:00 plus the equivalent times in the afternoon.


As suggested by Dorrulf and DqwertyC there are a minimum of 4 times when the digital clocks agree (if only (2|5), and (0|0) digits are considered to be mirror images, and a maximum of 11 times if (1|1) is also included.


Either way, 12 (or 13) trumps 11 (or 4), so the analog clock wins.

  • 3
    $\begingroup$ You need to include one midnight and one noon so 26 is correct. With 2 midnight, it would be 27! Also there are 2 extra matches at 6:00 and 18:00. $\endgroup$ – Phil1970 Oct 24 '18 at 1:21
  • $\begingroup$ Yes - I stand corrected. $\endgroup$ – Penguino Oct 24 '18 at 2:14

Analog clock matches are:

One match per ~55 minutes = 13 matches
An additional match at 6:00 = 1 match
Those matches occurs twice per day since the hour handle makes 2 complete turns by day.
(13 + 1) × 2 = 28 matches.

For the digital clock we have:

The following symmetries : 0 <--> 0, 1 <--> 1, 2 <--> 5 and 8 <--> 8
Starting at midnight, we have:
Leading 0: 00:00, 01:10, 02:50, 05:20 = 4 matches
Leading 1: 10:01, 11:11, 12:50, 15:20 = 4 matches
Leading 2: 20:05, 21:15, 22:55 = 3 matches
Leading 5: No matches
Leading 8: No matches
Thus we have 11 matches per day

So the result is:

The analog clock matches more often.


To me, the question asks particularly about when the values of the clocks read the same value on either wall, rather than just being exactly visually identical. This matters primarily for digital in that as long as the reflection shows actual numbers, and the value of said numbers is equal to that of the left wall, I count it. Essentially, the padding of the 1 doesn't matter, thus 11:11 is acceptable.

I believe the analog clock can be defined as:
f(y) = +-|x| for -3 <= x <= 3 where x is an integer.
The result is 13 points. Now, if you think that 12:30 should qualify and that the hour hand isn't off kilter enough to dissuade you, then you should also accept that the -0 and +0 y values in the function are different points, which would bump the results to 14. Given that this is only for a 12 hour period, we can double those numbers resulting in either 26 or 28 total values.

I haven't figured out how to represent the dialog clock as a function like above, but the bounds are as follows:
The clock value must read the same forward and back (palindrome).
The right side is limited to values between 0 and 24 (military time).
The left side is limited to values between 0 and 60.
Basically, the right side is the only one that matters, and the left side's tens position restrictions fall to the right's ones position, and the left's ones position restrictions fall to the right's tens position.
Therefore, we have values where: z = x*10 + y
Where 0 <= z <= 24 and 0 <= x <= 2 and 0 <= y <= 5
As 3, 4, 6, 7, 9 are not reflective, and 8 is too large.
This boils down to 11 results.

As such, analog has more matches. (As @Phil1970 already noted)

Original below. Can ignore.

As a forward, I'm going more for the idea that the clocks read identical times rather than they just look similar. The only analog positions I see matching would be at 12:30 and 3:45, and their self-reflections (6:00 and 9:15) for a total of 8 occurrences in 24-hr a day. (Thanks @ExcitedRaichu)
The digital clock has a few things of note beforehand:
The leading 0's and set to military time.
Again, if we function with the idea that as long as the times read the same (and the positions don't have to be exact - left oriented 1 versus right oriented 1), then we have these reflective cases: 0 to 0, 1 to 1, 2 to 5, 5 to 2. We can't use 8 because minutes only goes to 60. This gives us the following combinations (barring any were missed): 01:10, 11:11, 10:01, 15:21, 12:51, 05:20, 02:50. @DqwertC also found: 00:00, 02:50, 20:05, and 22:55.
That puts us at a ration of 8:11, in favor of the digital clock.

  • $\begingroup$ I'm still unsure of my understanding of "reflective agreement" though xD $\endgroup$ – Dorrulf Oct 23 '18 at 23:16
  • $\begingroup$ doesn’t (rot13) fvkbpybpx work for analog as well? $\endgroup$ – Excited Raichu Oct 23 '18 at 23:19
  • $\begingroup$ Also, most digital clocks don’t have an 0 in front of the hour number if it’s one digit $\endgroup$ – Excited Raichu Oct 23 '18 at 23:20
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    $\begingroup$ @ExcitedRaichu It explicitly says they do in the problem. $\endgroup$ – DqwertyC Oct 23 '18 at 23:21
  • $\begingroup$ @DqwertC whoops. missed that. $\endgroup$ – Excited Raichu Oct 23 '18 at 23:22

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