Alice and Bob were supposed to have a date at some o'clock. But, Bob was late for the date.

Alice was upset. She said "See you tomorrow at the same time." and went away.

Several interesting properties for their arrival times:

  1. Bob's arrival time is represented by three distinct digits.
  2. Rearranging those three digits, we get only one time before the appointment time and she arrived at that time.
  3. Bob was late within one hour.
  4. When we rearrange the three digits, other times are late more than one hour but less than two hours.

What time o'clock was their date?

  • $\begingroup$ For part 4, are all other times late more than one hour but less than two hours, or just some of them? Also, for part 2, is the appointment time the time they were supposed to meet, or is it Bob's arrival time? $\endgroup$ – Bailey M Aug 6 '15 at 14:50
  • $\begingroup$ @BaileyM We can make several times by permuting those digits. One is Alice's arrival time, one is Bob's arrival time, and all others satisfy the condition of part 4. The appointment time mentioned in part 2 is the time they were supposed to meet. $\endgroup$ – P.-S. Park Aug 6 '15 at 15:11
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    $\begingroup$ Looks like the problem's already been solved, but thanks either way, that does clear it up. :) $\endgroup$ – Bailey M Aug 6 '15 at 15:12
  • $\begingroup$ After reading the first sentence, I expected Eve to show up to the date in Bob's place. $\endgroup$ – Brian J Aug 6 '15 at 17:58
  • $\begingroup$ Does what time o'clock mean that the date is on the hour? Is there a single solution, or multiple? Does every rearrangement of the digits have to yield a proper time? $\endgroup$ – Cain Aug 6 '15 at 18:52

Letting Bob's arrival time be $x:yz$, we can infer:

According to 3:

The date was supposed to be at $x$:00.

According to 2:

We can get only one time before $x$:00, which is only possible if the smallest one of $x,y,z$ has less than two options for the minute (with at least one of the others > 5).

According to 4:

The other permutations are late by between one hour and two hours.


If there are two digits greater than 5, there'll only be 2 legal times, while there have to be at least 3 legal permutations. So there's only one such digit. Let's sort them all as $a<b<c>5$. Then out of $a:bc, b:ac, c:ab, c:ba$, only the first one is early, the second one is Bob's arrival time (appointment is at $b$:00), and the last two are late by between one hour and two hours. $c = b+1$, $b=5$, $c=6$, so they were supposed to meet at 5 o'clock. $a$ doesn't seem to matter, but 4:56 seems to be the most reasonable arrival time for Alice (then Bob must have arrived at 5:46).


Bob arrived at:


The date was supposed to be at:


The two times that are late by more than one hour, but less than two:

6:45 and 6:54

My thought process:

The three numbers had to be consecutive. Because each of them would be the hour digit for one of the given times, a non-consecutive triplet (i.e. 4,5,7) would give a latest and earliest time that were more than 2 hours apart. The lowest number must be greater than or equal to 6, so that there will only be 4 valid times (4:56 is valid; 4:65 is not). The two upper numbers must be less than 6, for the opposite reason, so 4,5,6 is the only triplet that can satisfy.

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    $\begingroup$ Ninja'd again! Nicely done. $\endgroup$ – Aggie Kidd Aug 6 '15 at 15:10
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    $\begingroup$ Haha, thank you! I am slightly embarrassed that it took your answer to make me realize that the date was scheduled for 5:00, and that Alice was in fact a few minutes early...! $\endgroup$ – Patrick N Aug 6 '15 at 15:20
  • $\begingroup$ How about 3:56? If they were supposed to meet at 5:00, Bob arrived at 5:36 and he was late within one hour after 5:00. $\endgroup$ – P.-S. Park Aug 6 '15 at 15:27
  • $\begingroup$ Looking at it, that satisfies the letter of the requirements, but it would mean that Alice was more than an hour early for their date, which seems a little excessive. $\endgroup$ – Patrick N Aug 6 '15 at 15:32
  • $\begingroup$ @Patrick Right, perhaps she loves him very much. :-) Anyway, however deeply she loves him, the answer is uniquely determined. $\endgroup$ – P.-S. Park Aug 6 '15 at 15:44

Bob arrived at


And just guessing they were supposed to meet at:

5 o'clock

The only earlier time then can be:

4:56, since 4:65 is not a valid time, and 4:56 is within one hour of 5:46. Alice is so good and arrived early.

The other arrangements are

6:54 and 6:45, both of which are more than an hour after 5 and within 2 hours of 5.

  • $\begingroup$ Ouch, Aggie. you keep getting the answer stolen :S $\endgroup$ – Moose Aug 6 '15 at 15:46
  • $\begingroup$ @Moose Gotta up my typing speed. $\endgroup$ – Aggie Kidd Aug 6 '15 at 15:47
  • $\begingroup$ either that or your reaction time / thinking speed. $\endgroup$ – Moose Aug 6 '15 at 15:47
  • $\begingroup$ @Moose Those could always use improvement. $\endgroup$ – Aggie Kidd Aug 6 '15 at 15:48
  • $\begingroup$ I think you're just outclassed by the sheer brainpower that is contained by the community of Puzzling SE $\endgroup$ – Moose Aug 6 '15 at 15:48

Alternate Answer

I don't know if this answer is just another possible solution, a proof that we are missing something (Since OP said the answer is uniquely determined), or just a matter of semantic disagreement.

Bob arrived at


Alice arrived at


Assuming that times of 0:xx are not valid, then the remaining times are

3:01 and 3:10

This lets the appointment time be

Anywhere between 1:11 and 1:29

Note: This seems to work with any two digits a and b such that 0 < a,b < 6 and a - b = 2

  • $\begingroup$ The question explicitly asks "What time o'clock was their date?". $\endgroup$ – Nautilus Aug 6 '15 at 16:43
  • $\begingroup$ Ooh, meaning the date had to be on the hour? That makes a lot more sense $\endgroup$ – Cain Aug 6 '15 at 16:44
  • $\begingroup$ @Nautilus I don't know that "what time o'clock" implies that it was specifically at some hour and zero minutes; o'clock just means "of the clock", and 1:11 is certainly a time of the clock. I think "at what hour was their date" would be totally clear. $\endgroup$ – amalloy Aug 6 '15 at 18:32
  • $\begingroup$ I don't buy it, as that would allow for way too many answers. $\endgroup$ – Nautilus Aug 6 '15 at 18:54
  • $\begingroup$ Oh I agree that's what was intended by the puzzle poser (because that interpretation leads to a unique answer), I'm just saying that "what time o'clock" is such a bad way to specify this constraint that it is totally reasonable to have missed the constraint entirely. $\endgroup$ – amalloy Aug 7 '15 at 0:31

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