6. There are 20 chairs arranged in a circle. There are $n$ people sitting in $n$ different chairs. These $n$ people stand, move $k$ chairs clockwise, and then sit again. After this happens, exactly the same set of chairs is occupied. (For example, if the 2nd, 4th, 7th, 9th, 12th, 14th, 17th, and 19th chairs are occupied to begin with, then exactly the same set of $n=8$ chairs is occupied after each person moves $k=15$ chairs clockwise.) For how many pairs $(n, k)$ with $1 \leq n \leq 20$ and $1 \leq k \leq 20$ is this possible?
6. We know that $1 \leq n \leq 20$ and $1 \leq k \leq 20$.
We focus on the possible values of $k$.
Consider $k=20$.
Since there are 20 chairs, then moving 20 chairs around the circle moves each person back to their original seat.
Therefore, any configuration of people sitting in chairs is preserved by this movement.
In other words, $n$ can take any value from 1 to 20 , inclusive, when $k=20$.
Thus, there are 20 pairs $(n, k)$ that work when $k=20$.
Consider $k=10$.
Here, any people in chairs 1 to 10 move to chairs 11 to 20 and any people in chairs 11 to 20 move to chairs 1 to 10 .
This means that the two halves ( 1 to 10 and 11 to 20) must contain the same number of people in the same configuration. (That is, if there are people in chairs 1, 3, 4, 8, then there are people in chairs $11,13,14,18$, and vice versa.) This means that the total number of people in chairs is even.
There are 10 possibilities for the number of chairs occupied among the first 10 chairs ( 1 to 10), and so 10 possibilities for the total number of chairs occupied (the even numbers from 2 to 20 , inclusive).
Thus, there are 10 pairs $(n, k)$ that work when $k=10$.
Consider $k=5$.
Here, any people in chairs 1 to 5 move to chairs 6 to 10, any people in chairs 6 to 10 move to chairs 11 to 15 , any people in chairs 11 to 15 move to chairs 16 to 20 , and any people in chairs 16 to 20 move to chairs 1 to 5 .
This means that the four sections containing 5 chairs must contain the same number of people in the same configuration.
This means that the total number of people in chairs is a multiple of 4 and can be $4,8,12,16$, or 20 , which gives 5 possible values of $n$.
Thus, there are 5 pairs $(n, k)$ that work when $k=5$.
If we consider $k=4$ and $k=2$, in a similar way, we can determine that there are 4 and 2 possible values for $n$, respectively.
Thus, there are 4 pairs $(n, k)$ that work when $k=4$ (namely, $(5,4),(10,4),(15,4),(20,4))$ and 2 pairs $(n, k)$ that work when $k=2$ (namely $(10,2)$ and $(20,2)$ ).
Consider $k=1$.
Here, each person in a chair moves 1 chair along the circle.
If there is a person in chair 1 , then they move to chair 2 , which means that there must have been a person in chair 2 . This person has moved to chair 3 , which means that there must have been a person in chair 3 , and so on.
Continuing, we see that all 20 chairs must be full, and so $n=20$ is the only possibility.
Thus, there is 1 pair $(n, k)$ that works when $k=1$.
Consider $k=3, k=7$ and $k=9$.
In each of these cases, $n=20$ is the only possibility.
If $k=3$, following a similar argument to that for $k=1$, the person in chair 1 moves to chair 4 , the person in chair 4 moves to chair 7 , and so on:
$$
\begin{aligned}
1 \rightarrow 4 & \rightarrow 7 \rightarrow 10 \rightarrow 13 \rightarrow 16 \rightarrow 19 \rightarrow 2 \rightarrow 5 \rightarrow 8 \rightarrow 11 \rightarrow 14 \\
& \rightarrow 17 \rightarrow 20 \rightarrow 3 \rightarrow 6 \rightarrow 9 \rightarrow 12 \rightarrow 15 \rightarrow 18 \rightarrow 1
\end{aligned}
$$

The cycle only concludes when all 20 chairs are included, and so if 1 chair is occupied, then all chairs are occupied.
A similar cycle can be constructed for $k=7$ and $k=9$ and so in each case only $n=20$ works.
Consider $k=6$.
Looking at the cycles as in the previous case,
$$
\begin{array}{c}
1 \rightarrow 7 \rightarrow 13 \rightarrow 19 \rightarrow 5 \rightarrow 11 \rightarrow 17 \rightarrow 3 \rightarrow 9 \rightarrow 15 \rightarrow 1 \\
2 \rightarrow 8 \rightarrow 14 \rightarrow 20 \rightarrow 6 \rightarrow 12 \rightarrow 18 \rightarrow 4 \rightarrow 10 \rightarrow 16 \rightarrow 2
\end{array}
$$

Note that starting at any chair other than 1 or 2 gives one of these two cycles. That is, all 20 chairs appear here.
This means that each set of 10 chairs are either all occupied or are all not occupied.
This means that there are two possible values of $n$, namely 10 or 20 .
Thus, there are 2 pairs $(n, k)$ that work when $k=6$.
Consider $k=8$.
Here, there are 4 cycles:
$$
\begin{array}{ll}
1 \rightarrow 9 \rightarrow 17 \rightarrow 5 \rightarrow 13 \rightarrow 1 & 2 \rightarrow 10 \rightarrow 18 \rightarrow 6 \rightarrow 14 \rightarrow 2 \\
3 \rightarrow 11 \rightarrow 19 \rightarrow 7 \rightarrow 15 \rightarrow 3 & 4 \rightarrow 12 \rightarrow 20 \rightarrow 8 \rightarrow 16 \rightarrow 4
\end{array}
$$
and so there are 4 values of $n$.
Note that starting at any chair other than $1,2,3$ or 4 gives one of these four cycles. That is, all 20 chairs appear here.
Thus, there are 4 pairs $(n, k)$ that work when $k=8$.
Finally, we consider $k=11,12,13,14,15,16,17,18,19$.
Moving 11 seats clockwise gives the same result as moving 9 seats counterclockwise. Since the arguments above do not depend on the actual direction, then the number of pairs when $k=11$ equals the number of pairs when $k=9$.
Similarly, the numbers of pairs for $k=12,13,14,15,16,17,18,19$ equal the number of pairs for $k=8,7,6,5,4,3,2,1$, respectively
This gives the following chart:
\begin{tabular}{c|c} 
Values of $k$ & Number of values of $n$ \\
\hline 20 & 20 \\
10 & 10 \\
5,15 & 5 \\
$4,8,12,16$ & 4 \\
$2,6,14,18$ & 2 \\
$1,3,7,9,11,13,17,19$ & 1
\end{tabular}

Therefore, the number of pairs $(n, k)$ is $20+10+2 \times 5+4 \times 4+4 \times 2+8 \times 1=72$.
ANSWER: 72