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.)
Solution i think i got it :) | Comment 10 of 13 |
To avoid making a mess with numbers, I will replace a few of the numbers with letters. Hand A will represent the six minute hand, Hand B will represent the four minute hand, and Hand C will be the 3 minute hand.

By finding a common factor, I can deduce that Hand A will make but only two revolutions around the clock. Every 12 minutes, the hand positions should fall back into their original place. Finding Hand A first would be most logical, since there are only two possible cases of revolutions it can be in.

First I found the time for hand A to skip an interval (from (1) to (2) is a single interval). The time is ( 6 min * 60 seconds ) / 12 intervals = 30 seconds.

Knowing this, I proceeded to find the distance (in seconds) of time first between (1) with (6), (10), and (11). I end up with:

[first column is revolution 1, second is revolution 2]

a->x | 150 | 510
a->y | 270 | 630
a->z | 300 | 660

You can instantly rule out the top four, why? Each of the hands fall perfectly into a number, so the seconds must be factorable by each hand.

The possibilities that remain are 300 and 660. Now I was lucky to choose 300 seconds first. If (1) travels to (11) in 300 seconds, one hand must travel 15 revolutions, and the other will travel 20. Remove 12 revolutions from each of these and you're left with 3 and 8 revolutions respectively.

Plug these in and sure enough it fits. (2) will travel (10) in 8 revolutions and (3) will travel to (6) in 3 revolutions.

(1)(11) = 6 minute hand
(2)(10) = 4 minute hand
(3)(6) = 3 minute hand

PS I beat turtles. ;)
  Posted by evan on 2004-02-21 03:08:57
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information