For some integers $n$, the expression $\frac{8(n-1)}{(n-1)(n-2)}$ is equal to an integer $M$. What is the sum of all possible values of $M$ ?
The expression $\frac{8(n-1)}{(n-1)(n-2)}$ is undefined when $n=1$. For every integer $n \neq 1$, the expression is equal to $\frac{8}{n-2}$.

Thus, if $n$ has the property that $\frac{8(n-1)}{(n-1)(n-2)}$ is also an integer, then $n$ must be an integer different from 1 with the property that $n-2$ is a factor of 8 .

The factors of 8 are

$$
-8,-4,-2,-1,1,2,4,8
$$

The following table has these values of $n-2$ in the first column, the corresponding values of $n$ in the second column, and the corresponding values of $\frac{8(n-1)}{(n-1)(n-2)}$ in the third column.

| $n-2$ | $n$ | $\frac{8(n-1)}{(n-1)(n-2)}$ |
| :---: | :---: | :---: |
| -8 | -6 | -1 |
| -4 | -2 | -2 |
| -2 | 0 | -4 |
| -1 | 1 | undefined |
| 1 | 3 | 8 |
| 2 | 4 | 4 |
| 4 | 6 | 2 |
| 8 | 10 | 1 |

Other than when $n=1$, we get an integer value of $\frac{8(n-1)}{(n-1)(n-2)}$.

The sum of the integer values of $\frac{8(n-1)}{(n-1)(n-2)}$ is $-1-2-4+8+4+2+1=8$.

ANSWER: 8