Two points are 5.2Km apart. Two people are initially at the opposite ends. They start running between the points back and forth with speeds of 13Kmph and 26Kmph respectively. They do this for 6 hours. How many times do they meet each other?

My attempt:

Position of 1st person after time $t=13t mod 5.2$

Position of 2nd person after time $t=5.2-26 mod 5.2$

So $13t mod 5.2=5.2-26t mod 5.2$

$13t mod 5.2+26t mod 5.2=5.2$

Taking mod on both sides

$(13t mod 5.2+26t mod 5.2)mod 5.2=5.2 mod 5.2$

$(13t+26t) mod 5.2=0$



Since $t<6 hours$


$k<45$ So they meet 45 times.

However, the options are 1, 2, 3 or 4 times

EDIT- I think my equations for their positions are wrong. How do I get the correct equations with explanation?

  • $\begingroup$ A quick comment: when doing modular arithmetic get rid of decimal points. Instead of working with KM as unit, use metres. $\endgroup$ – P Vanchinathan Dec 19 '18 at 2:42
  • $\begingroup$ @PVanchinathan But why? $\endgroup$ – Ryder Rude Dec 19 '18 at 2:45
  • $\begingroup$ Because modular arithmetic is about whole numbers and divisibility relations them and about remainders upon division. If you work with real numbers any number is divisible by any non-zero number and divisibility is not an issue at all. One does not talk of right boat design for fish, for humans yes. $\endgroup$ – P Vanchinathan Dec 19 '18 at 3:02
  • $\begingroup$ @PVanchinathan But my calculator works fine with decimals too. For example, 2pi mod 2.5 can be 2pi-5. Because 5 is the largest integral multiple of 2.5 which is smaller than 2pi. I don't think there's any problem with defining mod like this. It's just the k-values which must be integers for this to work. $\endgroup$ – Ryder Rude Dec 19 '18 at 3:08
  • $\begingroup$ Here you are using $\pi$ with a unit, radians, what physicists call quantity with dimension, something that measures angles. One can use an alternative unit and work with integral multiples of $\pi$. In number theory WHOLE numbers are dimensionless objects. $\endgroup$ – P Vanchinathan Dec 19 '18 at 3:11

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