Problem about clocks that gain $2$, $6$ and $12$ minutes per hour respectively Three old clocks, whose hour hands are missing, have minute hands which all run fast. Clock P, Q and R gain 2, 6 and 12 minutes per hour respectively. They start at midday with all three minutes hands pointing to 12. Find the number of hours late when all three hands show the same number of minutes? 
An olympiad student asked me this problem but I am struggling to find an elementary solution
 A: The first minute hand ticks $62$ times in an hour, and the second one ticks $66$ times in an hour. The difference between the first two clocks is therefore $4$ ticks per hour. How long does it take before this difference is a multiple of $60$ ticks (which is what it means that they show the same)? Do the same between the second and third.
So now you know that the first two clocks align every $x$ hours, and the last two clocks align every $y$ hours. How often does this mean that the two alignments... align?
A: Let $h$ be the required number of hours. 

Then $h$ is the least positive real number such that, mod $60$, we have
$$62h \equiv 66h \equiv 72h$$
Restated in terms of divisibility, we get
$$60|4h\;\;\text{and}\;\;60|6h$$
which simplifies to
$$15|h\;\;\text{and}\;\;10|h$$
Hence, since we want the least $h$, it follows that $h$ is the LCM of $10$ and $15$, so $h = 30$.

Notes:


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*If $x,y,z$ are real numbers, the notation $x \equiv y\;(\text{mod}\;z)$ means $x-y = nz$, for some integer $n$.$\\[2pt]$

*If $x,y$ are real numbers, the notation $x|y$ means $y=nx$, for some integer $n$.

