Question 1098484
3 clocks, hour hands are missing, have minute hands running fast.

Clock P, Q, R gain 2, 6, 12 mins per hour respectively.

They start at midday - pointing to 12.

Find the number of hours later when all 3 hands show the same number of minutes.
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They make 1 revolution in 58, 54 and 48 minutes.
The smallest number evenly divisible by the 3 times is
29*27*16 = 12528 minutes
= 208 hours 48 minutes
= 8 days, 16 hours, 48 minutes
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The 3 minute hands will again coincide at 12.
Corrected after I realized I missed a factor of 3.
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There might be another conjunction sooner at some other position.
I'm thinking about that.
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Remember, a stopped clock is right twice each day.
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Interesting problem.
The only coincident position is the 12 o'clock position, which is not obvious at first.
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30 hours is obviously correct.  Actually, every 30 hours.