I looked at my watch at 12:05. I wasn't sure, for a moment, whether it was 12:05 or 1:00; I had to carefully look to determine which of the two hands was the longer one.
A question for you: how many times in a given twelve-hour period could I have this problem? More rigorously, suppose I have an ordinary twelve-hour analog clock, with an hour hand and a minute hand but no second hand. Furthermore suppose I can measure the position of the hands absolutely precisely, and they're "sweep" hands (i. e. they move at a constant angular rate, without "ticks"). At how many times between (say) noon and midnight could I interchange the hands of the clock and still have the hands in a position that corresponds to some time -- but not the time that it actually is? Noon, for example, is not such a time; if I interchange the minute and hour hands at noon I get a valid position of the hands, but that's the position the corresponds to noon. (I won't give an example of a valid time because giving one would be a big hint.)
Bonus: what are these times?
Another bonus: Add a second hand; are there still times which give rise to ambiguous hand configurations? (I don't know the answer to this one.)
(No fair looking up a solution; this is actually a pretty well-known brainteaser. It's well-known enough that I probably knew it existed, somewhere in the back of my mind, before I reinvented it today.)
edit (1:14 pm): Boris points out that he wrote a very similar question as question 23 of this test (PDF).
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11 comments:
Without busting my brain too hard, I get 48 times (in 24 hrs). What those times are I have yet to figure. Not even sure if the number of events is correct.
I wrote essentially this problem for the 2004 UGA High School Math Tournament. It's problem 23 on the written test. It doesn't exclude times where the hands are in the same place, but that's an easy modification to the solution given there.
More problems solvable analogously: Problem 14 from 2005 asks how many times during the day the hands are directly opposite each other. Problem 20 from 2003 asks how many times they are perpendicular.
132 times in 12 hours? Or am I way off track here?
Hot Cup, that's the right answer.
Yah, I thought about that right after lunch. It makes sense.
> At how many times between (say) noon and midnight could I interchange the hands of the clock and still have the hands in a position that corresponds to some time -- but not the time that it actually is?
An infinite number of times, of course. Unless you claim that a period of time is composed of a finite number of moments.
It's likely that you mean to limit the question to distinct (hour, minute) pairs though.
/pedantry I just couldn't hold back
Dang! I had recently been thinking about posing this problem as a POW (problem of the week) on our blog. Good thing, I waited!
@Bill Mill: Your pedantry fails because Michael specified that the hands sweep, rather than tick, and that you can measure them precisely.
@Michael: I computed over lunch that there are no such times if you also have a second hand. I'm fairly confident that I didn't mess up.
I agree with Boris that the second hand will make all of the times unique. I haven't shown it to myself exhaustively or generally, but I'm fairly satisfied...unless you meant that you could confuse the second hand for another hand. I haven't even tried to work that out. (But the second hand is harder to confuse for another hand, anyway.)
Turns out that if you have 3 hands, you can never switch any two, or all three in any order, to obtain a valid time (except at 12:00). A friend and I calculated this once.
Pete Winkler has this puzzle in his book (of puzzles), as far as I remember. This problem is essentially topological (the arms can progress in a non-linear fashion); the answer is (12,1)x(1,12)-(12,1)x(1,1)=143-11=132 (here x is the vector product, (a,b)x(c,d)=ad-bc).
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