Posted June 5, 2014 It is 12MN when all the hands of this unusual watch rotates clockwise at proper speeds. But all the hands runs in opposite direction when they strike 12..like a pendulum. What is the real time if it reads false as shown below ? 1 Share this post Link to post Share on other sites
0 Posted June 7, 2014 Interesting puzzle, and more amenable to solving than it first seems. Label the hands H, M and S. Every 2 minutes S returns to initial position and handedness. Every 2 hours M returns to initial position and handedness. Every day H returns to initial position and handedness. So we can't tell how many days have elapsed from the start. But time of day is all that is asked, so the puzzle should have a solution. Initially, and after any even number of minutes S turns CW (and vv.) Initially, and after any even number of hours, M turns CW (and vv.) Before noon, H turns CW; after noon H turns CCW. So there are two cases for the time = hh:mm:ss. Before noon.H turns CW. hh=08 (even). M turns CW => mm=32 (even). S turns CW => ss=26.Time = 08:32:26 After noon.H turns CCW. h=15 (odd). M turns CCW => mm=27 (odd). S turns CCW => ss=34.Time = 15:27:34 Which is it - all hands moving CW (08:32:26) or all moving CCW (15:27:34)? The only clues I can find are the precise positions of H and M. If M and S are turning CW, M would be closer to 6 than to 7 b/c S hasn't got to 6 yet.If M and S are CCW, M would be closer to 6 than to 7 b/c S has passed 6. Either way, M should be slightly closer to 6 than to 7.M appears to be slightly closer to 7. Contradiction with either case. If H and M are turning CW, H would be closer to 9 than to 8 b/c M has passed 6. If H and M are CCW, H would be closer to 9 than to 8 b/c M hasn't got to 6 yet.Either way, H should be slightly closer to 9 than to 8.H appears to be slightly closer to 9. Consistent with either case. I'm out of clues: the time is ambiguous. Thanks for noticing Bonanova. I see that point clearly . With Excel I tried to pull the green line nearer to next mark (nearer to six) but the it always snaps back to center. However, your solution is the solution.. When we add the AM time with its same position on PM time it is 12:00:00.. twice on a position per day. 15:27:34 If the time is 08:32:26 AM it is true time. 0 Share this post Link to post Share on other sites
0 Posted June 6, 2014 Is it correct to conclude The hands begin rotating CW, all pointing at 12, at midnight Each hand reverses direction each time it approaches 12, from either direction The clock may have been running for an arbitrarily long time when the photograph was taken 0 Share this post Link to post Share on other sites
0 Posted June 6, 2014 Is it correct to conclude The hands begin rotating CW, all pointing at 12, at midnight Each hand reverses direction each time it approaches 12, from either direction The clock may have been running for an arbitrarily long time when the photograph was taken Correct conclusions.. 0 Share this post Link to post Share on other sites
0 Posted June 7, 2014 Interesting puzzle, and more amenable to solving than it first seems. Label the hands H, M and S. Every 2 minutes S returns to initial position and handedness. Every 2 hours M returns to initial position and handedness. Every day H returns to initial position and handedness. So we can't tell how many days have elapsed from the start. But time of day is all that is asked, so the puzzle should have a solution. Initially, and after any even number of minutes S turns CW (and vv.) Initially, and after any even number of hours, M turns CW (and vv.) Before noon, H turns CW; after noon H turns CCW. So there are two cases for the time = hh:mm:ss. Before noon.H turns CW. hh=08 (even). M turns CW => mm=32 (even). S turns CW => ss=26.Time = 08:32:26After noon.H turns CCW. h=15 (odd). M turns CCW => mm=27 (odd). S turns CCW => ss=34.Time = 15:27:34 Which is it - all hands moving CW (08:32:26) or all moving CCW (15:27:34)? The only clues I can find are the precise positions of H and M. If M and S are turning CW, M would be closer to 6 than to 7 b/c S hasn't got to 6 yet.If M and S are CCW, M would be closer to 6 than to 7 b/c S has passed 6. Either way, M should be slightly closer to 6 than to 7.M appears to be slightly closer to 7. Contradiction with either case.If H and M are turning CW, H would be closer to 9 than to 8 b/c M has passed 6. If H and M are CCW, H would be closer to 9 than to 8 b/c M hasn't got to 6 yet.Either way, H should be slightly closer to 9 than to 8.H appears to be slightly closer to 9. Consistent with either case. I'm out of clues: the time is ambiguous. 1 Share this post Link to post Share on other sites
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