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September 3rd, 2018, 02:43 AM  #11 
Global Moderator Joined: Dec 2006 Posts: 19,882 Thanks: 1835 
My last example was of that type.

September 3rd, 2018, 10:03 AM  #12 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
Assume after k,m,n laps A,B,C meet at a point a on the track:at time t: 3t=100k+a 5t=100m+a 8t=100n+a (100k+a)/3=(100m+a)/5=(100n+a)/8 From first equality, 500k+5a=300m+3a, 2a=300m500k From second equality 3a=500n800m $\displaystyle \rightarrow$ (200m500k)/2=(500n800m)/3 $\displaystyle \rightarrow$ 25m15k10n=0, n>m>k, and by inspection k=2, m=4, n=7. Then a=100 (1 lap) and A,B,C meet after running 3,5,8 laps respectively, ie, they meet at starting point. Last edited by zylo; September 3rd, 2018 at 10:28 AM. Reason: add "they meet at starting point," typo 
September 3rd, 2018, 02:01 PM  #13 
Global Moderator Joined: Dec 2006 Posts: 19,882 Thanks: 1835 
For the original problem, one can use a = 0, t = 100, and (k, m, n) = (3, 5, .


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circular, problem, solve, track 
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