Question on Clocks 
Three watches are set together. The first gains $5$ minutes a week, the second gains $8$ minutes a week, whilst the third loses $4$ minutes a week. When will they again indicate the same time?

Please give me some hint on how to tackle this problem.
 A: Assumptions:


*

*12-hour watches.  So two watches that differ by $12 \times 60\ $ accumulated minutes are indicating the same time.  So not one of these.

*"The same time" means "have hands pointing at the same labels" (no a.m. vs. p.m. tracking, so not one of these)

*Time is (miraculously) indicated with infinite precision.  (The accuracy is somewhat lacking, though.)


The first two differ by 3 minutes per week, so at least $\frac{12 \times 60}{3} = 240$ weeks must pass before the first two watches indicate the same time.  This will recur every 240 weeks.  The first and last differ by 9 minutes per week, so $\frac{12 \times 60}{9} = 80$ weeks is the time between matches of the first and last watch.  Conveniently, this divides 240 weeks, so all three watches will agree every 240 weeks.
A: There are $720$ minutes in $12$ hours. So, the time on the watches is modulo $720$ minutes. 
Thus, the clocks show the same time after $n$ weeks iff $5n \equiv 8n \equiv -4n \pmod{720}$. 
This condition is met iff $3n \equiv 0 \pmod{720}$, i.e. $n$ is a multiple of $240$. 
So, the watches will show the same time after $240$ weeks. 
