# 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 $B$?

• you might consider that any such equation (for the various values of N) are some curves in the A-B-plane. So you could plot those for the different values of N on a quad paper. So you could "see" the possible solutions... – Dr. Richard Klitzing May 8 '18 at 21:09
• I don’t really understand what you mean @Dr.RichardKlitzing – Ali Mosallaei May 8 '18 at 21:10
• A and B are 2 variables. An equation in those variables will lead to graph in the A-B-plane. So you'd plot curve $C_0 = \{(A,B) | 0=A*B\}$, curve $C_1 = \{(A,B) | A+B=A*B\}$, $C_2 = \{(A,B) | 2*(A+B)=A*B\}$, etc. - You then just have to look where the cartesian squares of the graph paper (integer solutions!) intersect those graphs... – Dr. Richard Klitzing May 8 '18 at 21:17

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 $n=4$, $$B = \frac{4A}{A-4} \implies (A,B)\in \{(8,8)\}.$$

• Awesome! Thanks! – Ali Mosallaei May 8 '18 at 21:33
• While your answer is very easy to read, you forgot multiple of 10, as Robert points out in the other answer – SK19 May 8 '18 at 22:48
• I can't edit in the $0,$ on $n=1,2,3,4$ -_- – SK19 May 9 '18 at 1:31
• @SK19 Thanks. Updated the answer. – Math Lover May 9 '18 at 1:35

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 $10, 20, 22, 30, 36, 40, 44, 50, 60, 63, 66, 70, 80, 88, 90$.