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 $B$?
 A: 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 $n=4$,
$$B = \frac{4A}{A-4} \implies (A,B)\in \{(8,8)\}.$$
A: You want $d_1 \in \{1,\ldots, 9\}$ and $d_2 \in \{0,\ldots,9\}$ such that $ \dfrac{d_1 d_2}{d_1 + d_2}=k$ is an integer. Note that if $d_2 > d_1$, we get another solution by interchanging $d_1$ and $d_2$, so we may assume $d_2 \le d_1$.   With $s = d_1 + d_2$, our equation can be written as  $k = d_1 - d_1^2/s$. Thus $s$ is a factor of $d_1^2$, with $d_1 \le s \le 2 d_1$.  Now we have to consider some cases.


*

*If $d_1 = 1$, the only factor of $1^2$ is $1$, so $d_2 = 0$.

*If $d_1$ is a prime ($2, 3, 5, 7$) or prime power ($4, 8, 9$), the only factors $s$ of $d_1^2$ are powers of the same prime; this is in $[d_1, 2d_1]$ only if $s=d_1$ (leading to $d_2 = 0$) or (if the prime is $2$) $s = 2 d_1$ (leading to $d_1 = d_2=2$ or $4$ or $8$).

*The only other case is $d_1 = 6$: the possibilities are $s = 6, 9, 12$ corresponding to $d_2 = 0, 3, 6$.


Thus the possible numbers are $10, 20, 22, 30, 36, 40, 44, 50, 60, 63, 66, 70, 80, 88, 90$.
