# Question #de60e

Feb 14, 2017

Ken will have run $3$ laps, and Hamid will have run $4$.

#### Explanation:

In order for Ken and Hamid to meet again exactly on the start line, each must have completed a positive integer ("whole") number of laps, since if they met anywhere else, their current lap would be in progress.

Since time affects both equally, the same amount of time must have passed when they first meet. To find that time, simply take the time in which the slowest one (Ken) completes a lap, then see if the other time value divides it perfectly (no remainder, integer quotient). If not, then check the second, third, fourth, etc multiples of Ken's time, and find the first one that is divided by Hamid's time perfectly, or the first (lowest) multiple of $80$ that is divisible by $60$. This means we need to check $80 , 160 , 240 , 320 , 400 ,$ etc.

The reason we are doing this, is that if that number is divisible by $60$, Hamid will also have run a whole number of laps, so they will then both be at the start line. We want to pick the lowest of those numbers, to find the first time this happens.

Essentially, we are trying to find the least common factor of $80$ and $60$. We will find that this is $240$, since:

$240 \div 80 = 3$ and $240 \div 60 = 4$. Therefore,

Ken will have run $3$ laps, since he runs $1$ in $80$ seconds, it takes him $240$ seconds to run $3$. For the same reason, Hamid will have run $4$.