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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 my solution -no calculators | Comment 4 of 13 |
The hands advance: 4 numbers/min....denote hand A
3 numbers/min .... B
2 numbers/min ..... C
Only hand A can reach 10 by even number per minute in 2, 5, 8 , 11 etc increments of four,
derived from 2+4*k=10 mod 12
Now looking for the hour eleven it is either
1+2*k=11 or 3+3*k=11 (all mod 12)
3+3*k=11 is impossible since only one side is divisible by three so 3+3*k=6 mod 12 is the third equation,the second being 1+2*k=11 mod 12.
So far we have identified the hands :
A on 2 k=2,5,8,11,...
B on 3 k=1,5,9 ,13,
C on 1 k=5,11,17

ans; new position within 5 minutes:
2+5*4= 22==>10
3+5*3= 18==>6
1+5*2= 11
q.e.d.

ady
  Posted by Ady TZIDON on 2004-02-18 03:49:37
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