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.
(In reply to
re(2): Solution by Mital)
Mital:
"u saw
3 min / circle @ 2 & 11
4 min / circle @ 3 & 6
6 min / circle @ 1 & 10 "
ok
"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,"
Reply:
In the first part of the above sentence (before the "so") you say that 3, 6, 9 and 12 as positions for the 4min 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 4min hand advances 3, 6, 9 or 12 positions, the other hands are on exact positions.
Mital:
" 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."
Reply:
Again, this depends on assuming that the 3min hand starts on the 2. But even then it is incorrect. If the 3min hand starts on 2, then three minutes later the 6min hand will be on the 7 if it was on the 1 to begin with, while the 3min hand is back at 2.
Mital:
" & whenever 3 min hand on 11 6 min will be on 10. "
Reply:
If the 3min hand starts on the 2, it will take (112)*3/12 = 9/4 minutes to reach the 11. At that time, the 6min hand will have advanced only half as faror four and a half positions or if it goes once more around, 10 and a half positions either way, the 6min hand would not be at an exact integral position.

Posted by Charlie
on 20040218 15:31:10 