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 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.

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 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.

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

Reply:
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