Three runners called them Tom, Dick and Harry run along a circular route. Tom takes 5 minutes to complete a round, Dick takes 7 minutes, and Harry takes 11 minutes. Tom and Harry start their run at 7:00 am sharp, but Dick fills his water bottle first and starts at 7:01 am. After how many minutes, minimum, will they complete a round at exactly the same time?
I know I could compute this without too much difficulty by listing the times at which Dick and Harry complete a lap and finding their first shared multiple of 5 ( since Tom has a period of 5 and started at 7:00am, this would also be the first time he shared with both Dick and Henry).
However, I'm wondering if there is a more elegant way to find the solution, perhaps involving congruences?
When this event happens after $N$ minutes the three will have completed $t$, $d$, and $h$ rounds respectively, and one has the equations $$N=5t=7d+1=11h\ .$$ This at once implies $t=11x$ and $h=5x$ for some $x>0$, hence $$N=55x=7d+1\ .$$ In particular $55x\equiv1$ mod $7$, and as $55\equiv-1$ mod $7$ we necessarily have $x\equiv-1$ mod $7$ as well. Therefore the smallest admissible $x$ is $x=6$, so that we obtain $N=330$ [min] as solution of the problem.