Why do these days of the week line up?

Let's think of numbers $0, 1, \ldots, 6, \ldots$ as the days starting from some Sunday.

Then every Monday, Wednesday, and Friday, means $\{7n + 1, 7n + 3, 7n + 5\}$.

"Every three days" is all days of the form $3n + k$, where $k \in \{0, 1, 2\}$. Equivalently, these are all days that are $\equiv k \pmod 3$.

Taking $\{\rm M, W, F\}$ modulo $3$, we get $\{n+1, n, n+2\}$, which is the same as $\{0,1,2\}$, thus one of Monday, Wednesday, Friday is $k \pmod{3}$. Thus for any $k$, there is a day that matches both sets.

Take any three numbers of the form $n,n+2,n+4$ for some $n\in\Bbb N$. Then one of them, and exactly one, is a multiple of $3$.

This is because $2$ is a generator of the cyclic group of order $3$, $\Bbb Z/3\Bbb Z$.

It's clear the number of times a week I see him has an upper bound of 1. If I see him Mon, he'll come in Thurs and I won't see him Wed/Fri. The same logic can be used for if I see him Wed/Fri.

Similar logic works the other way around. Since he comes every three days, he'll have to be there on Monday, Tuesday or Wednesday:

• If he comes in on Monday, you'll see him on Monday.
• If he comes in on Tuesday, he'll also come on Friday, and you'll see him on Friday.
• If he comes in on Wednesday, you'll see him on Wednesday.

So there's also a lower bound of 1.

If you go on three days that are four days apart, they can be numbered as $n$, $n+4$, and $n+8$. Modulo 3, these are $n$, $n+1$ and $n+2$. All remainders modulo 3 are covered, so exactly one of these days coincides with the every-three-days cycle of the other person.

On the other hand, if you went on days that are three apart (or any multiple of three), then your partial cycle would be locked with the other person's cycle, and you'd see him either every time, or never.