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 September 1st, 2018, 04:03 AM #1 Senior Member   Joined: Aug 2014 From: India Posts: 343 Thanks: 1 How to solve this circular track problem? On a circular track of length 100 metres A,B and C start running clockwise from the same point, with speeds of 3m/sec, 5m/sec and 8m/sec respectively. At what distance from the starting position would they meet for the first time anywhere on the track?
 September 1st, 2018, 10:45 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,311 Thanks: 1223 basically you need to find $t$ such that $3 t = 5t = 8t \pmod{100}$ Note that $3,~5,~8$ have no factors in common. So it must be that $t=100k$ for this equality to hold. Obviously the first such occurrence other than at the start is $k=1$ which corresponds to $t=100s$ This corresponds to a distance of $0m$ from the starting line.
 September 1st, 2018, 01:36 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,958 Thanks: 991 Ganesh, did you at least google "circular track problems"?
September 1st, 2018, 02:29 PM   #4
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Quote:
 Originally Posted by Denis Ganesh, did you at least google "circular track problems"?
don't be absurd

September 2nd, 2018, 04:52 AM   #5
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Quote:
 Originally Posted by romsek basically you need to find $t$ such that $3 t = 5t = 8t \pmod{100}$
can't you proceed without using modular arithmetic?

 September 2nd, 2018, 07:45 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,285 Thanks: 1967 Every 20 seconds, C laps A. Every 50 seconds, B laps A. Hence A is first lapped at the same time by B and C after 100 seconds. The reasoning romsek gave wasn't quite right. For the equation 3t ≡ 5t ≡ 7t (mod 100), the coefficients would also have no common factor (other than 1), but the solution would then be t = 50, not t = 100. Thanks from romsek
September 2nd, 2018, 09:40 AM   #7
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Quote:
 Originally Posted by skipjack Every 20 seconds, C laps A. Every 50 seconds, B laps A. Hence A is first lapped at the same time by B and C after 100 seconds. The reasoning romsek gave wasn't quite right. For the equation 3t ≡ 5t ≡ 7t (mod 100), the coefficients would also have no common factor (other than 1), but the solution would then be t = 50, not t = 100.
3x50 = 150 % 100 = 50

5x50 = 250 % 100 = 50

8x50 = 400 % 100 = 0

I think you are mistaken.

 September 2nd, 2018, 11:58 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,285 Thanks: 1967 I had replaced 8 with 7 to show that you got the correct answer, but by an incorrect method. Thanks from romsek
 September 2nd, 2018, 12:16 PM #9 Newbie   Joined: Dec 2016 From: Austin Posts: 16 Thanks: 1 There are a few different ways of solving this. Could you let us know what type of course this question was asked in? For example, if this is for a Physics course, I might suggest an angular speed equation. If it is Geometry based, then a circumference approach. If it were a challenge problem in an introductory Algebra course, then we might use the Least Common Multiple and a couple of problem solving techniques. Thanks!
 September 2nd, 2018, 12:32 PM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,958 Thanks: 991 Can someone provide an example where the 3 runners meet not at the starting point? Thanks. I must have goofed in my program...

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