Number $AB$ if $A\times B=N(A+B)$? 
Number $36$ has the property that the sum of digits $3+6=9$ is a factor of
  the product of the digits $3\times 6=18$. Find all two-digit numbers with this property.

So I started by stating the obvious, that numbers $22$, $44$, $66$, and  $88$ can work. Then I made an equation:
$$(A+B)N=AB,$$
where $N$ can take on different values. I plugged in let’s say $N=1$ and get $A+B=AB$, but this doesn’t really give me any values without guessing and checking. Does anyone know how to solve without guessing and checking values of $A$ and $B$?

Note that $$(A+B)n=AB \implies B = \frac{nA}{A-n},$$ which is a decreasing function of $A$. Since the maximum value of $A$ is $9$, we have
$$\frac{9n}{9-n}\le 9 \implies n \le 4.$$
Therefore, we only need to consider $n=0, 1,2,3,4$. For them, we have the following $(A,B)$ pairs:
First, $n=0$ implies
$$B = 0 \implies (A,B) \in \{(1,0),(2,0),\cdots,(9,0)\}$$
For $n=1$,
$$B = \frac{A}{A-1} \implies (A,B)\in \{(2,2)\}$$
For $n=2$,
$$B = \frac{2A}{A-2} \implies (A,B)\in \{(3,6),(4,4),(6,3)\}.$$
For $n=3$,
$$B = \frac{3A}{A-3} \implies (A,B)\in \{(6,6)\}.$$
For $n=4$,
$$B = \frac{4A}{A-4} \implies (A,B)\in \{(8,8)\}.$$
