All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
The strange clock (Posted on 2004-02-17) Difficulty: 3 of 5
I have a very strange clock. At first glance, it looks like a normal clock with three hands and the numbers 1 through 12 all around. The only differences are that the hands are indistinguishable from each other and they are faster. One hand completes a circle in 3 minutes, another in 4 minutes, and the last in 6 minutes. They all go clockwise.

One morning, when I looked at the clock, the hands were all pointing exactly at the numbers 1, 2, and 3.
Later that day, I saw that the three hands were pointing exactly at 6, 10, and 11.

Can you identify which hands I saw each time? Prove it.

See The Solution Submitted by Tristan    
Rating: 2.6667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): Solution | Comment 8 of 13 |
(In reply to re(2): Solution by Mital)

"u saw
3 min / circle @ 2 & 11
4 min / circle @ 3 & 6
6 min / circle @ 1 & 10 "

"4 min hand rquires 20 secs to cover normal 5 sec distance so whenever 4 min hand goes 60,120,180 & 240 secs i.e. on 3 , 6 , 9 , 12 other two hands are on exatly on the number of clock, so 4 min hand will be at 3 & 6 because for any other position of that other two hands will be in between the to numbers,"

In the first part of the above sentence (before the "so") you say that 3, 6, 9 and 12 as positions for the 4-min hand all allow the other hands to be on exact positions, but in the latter part you restrict it to 3 and 6. Also, the absolute positions depend on where it started. The 3 and 6 are indeed valid if it starts on 3, but how did you know that? What you know at the beginning is that when the 4-min hand advances 3, 6, 9 or 12 positions, the other hands are on exact positions.

" now 3 min hand is twice as fast as 6 min , with each requires 15 sec & 30 secs for normal 1 min distance, so whenever 3 min hand is on two , 6 min will be at 1."

Again, this depends on assuming that the 3-min hand starts on the 2. But even then it is incorrect. If the 3-min hand starts on 2, then three minutes later the 6-min hand will be on the 7 if it was on the 1 to begin with, while the 3-min hand is back at 2.

" & whenever 3 min hand on 11 6 min will be on 10. "

If the 3-min hand starts on the 2, it will take (11-2)*3/12 = 9/4 minutes to reach the 11. At that time, the 6-min hand will have advanced only half as far--or four and a half positions-- or if it goes once more around, 10 and a half positions-- either way, the 6-min hand would not be at an exact integral position.

  Posted by Charlie on 2004-02-18 15:31:10

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information