# When will she pass me for the first time? [solution]

Recently, my dad posed the following question here:

My wife and I walk on a circular track, starting at the same point.  She does m laps in the time that it takes me to do n laps.  She walks faster than I do, so m > n.  After how many laps will she catch up with me again?

If you haven’t solved it yet, give it a crack. It’s a fun problem that has surprising depth.

Here’s my solution (in it, I refer to “mom” rather than “my wife” for obvious reasons!):

Since mom’s lap rate is $m$ laps per unit time, and dad’s lap rate is $n$ laps per unit time, in time $t$, mom goes $mt$ laps and dad goes $nt$ laps.

They meet whenever their distance (measured in laps) is separated by an integer number of laps $k$. That is, mom and dad meet when

$mt=nt+k, k\in\mathbb{Z}.$

This happens at time

$t=\frac{k}{m-n}.$

Mom will have gone

$mt=\frac{mk}{m-n}$

laps and dad will have gone

$nt=\frac{nk}{m-n}$

laps when they meet for the $k$th time.

And that’s it! That’s the general solution. This means that:

• At time $t=0$, dad and mom “meet” because they haven’t even started walking at all (they are $k=0$ laps apart).
• At time $t=\frac{1}{m-n}$, dad and mom meet for their first time after having started walking (they are $k=1$ lap apart). This is the answer to the problem as it was originally stated. Mom will have gone $mt=\frac{m}{m-n}$ laps and dad will have gone $nt=\frac{n}{m-n}$ laps when they meet for the first time.
• At time $t=\frac{2}{m-n}$, dad and mom meet for their second time (now $k=2$ laps apart).
• At time $t=\frac{k}{m-n}$, dad and mom meet for their $k$th time.

Here are two examples:

• If mom walks 15 laps in the time it takes dad to walk 10 laps, when they meet up for the first time, mom will have gone $\frac{m}{m-n}=3$ laps and dad will have gone $\frac{n}{m-n}=2$ laps.
• If mom walks 12 laps in the time it takes dad to walk 5 laps, when they meet up for the first time, mom will have gone $\frac{m}{m-n}=1\frac{5}{7}$ laps and dad will have gone $\frac{n}{m-n}=\frac{5}{7}$ laps.

Boom! Problem solved! 🙂