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August 15th, 2010, 12:05 PM   #1
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the clock hands

I don't know if it's the right place to ask this question so I'm sorry if it's not. I'm a rather layperson in mathematics but there's one puzzle I've been given by my friend recently and which i have to solve in order to borrow something from her. Can you help me with this, please? It goes like this:
"When I checked my watch this morning the hour hand was exactly where the minute hand is now, and the minute hand was one minute division before where the hour hand is now. And both hands are exactly at minute divisions now. What was the time when I checked it in the morning?"
I went like: x=the place of the hour hand now, y=the place of the minute hand now, likewise z=the place of the hour hand earlier, w=the place of the minute hand earlier
So: (60z + w)/12=y and (60x + y)/12 - 1=w
It made the final equation 720z + 60x - 143y = 12
What should I do next? I don't know how to do the equations with three unknowns (though the unknowns are all integers)
Anyone can please explain how I should proceed? I would appreciate that.
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August 15th, 2010, 09:34 PM   #2
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Re: the clock hands

7:12.

At 2:36 the hour hand is 13 ticks past 12 and the minute hand is 36 ticks past twelve. In the morning the minute hand was 12 ticks past 12 and the hour hand was 36 ticks past 12.
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August 16th, 2010, 01:52 AM   #3
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Re: the clock hands

Thanks a lot! I got what I wanted. How did you do that? you did any calculations or you just knew it straight away? I'm just curious...
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August 16th, 2010, 09:25 AM   #4
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Re: the clock hands

I wrote
p = b - 1
q = a
a
b

and

12
24
36
48
00

then the answer became apparent.
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August 16th, 2010, 10:47 AM   #5
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Re: the clock hands

Right...foolish me give me some of your brains Greg please...Thanks once again.
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