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September 3rd, 2018, 02:43 AM   #11
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My last example was of that type.
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September 3rd, 2018, 10:03 AM   #12
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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=300m-500k
From second equality
3a=500n-800m
$\displaystyle \rightarrow$
(200m-500k)/2=(500n-800m)/3
$\displaystyle \rightarrow$
25m-15k-10n=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
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September 3rd, 2018, 02:01 PM   #13
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For the original problem, one can use a = 0, t = 100, and (k, m, n) = (3, 5, .
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