**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 laps per unit time, and dad’s lap rate is laps per unit time, in time , mom goes laps and dad goes laps.

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

This happens at time

Mom will have gone

laps and dad will have gone

laps when they meet for the th time.

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

- At time , dad and mom “meet” because they haven’t even started walking at all (they are laps apart).
- At time , dad and mom meet for their first time after having started walking (they are lap apart).
**This is the answer to the problem as it was originally stated.**Mom will have gone laps and dad will have gone laps when they meet for the first time. - At time , dad and mom meet for their second time (now laps apart).
- At time , dad and mom meet for their 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 laps and dad will have gone 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 laps and dad will have gone laps.

Boom! Problem solved! 🙂