Digit Sum = Digit Product

The problem becomes a lot simpler when we start with finding all the solutions in the $1$ digit numbers, then progressing to $2$ digit numbers, $3$ digit and finally $4$ digit solutions.

For $1$ digit numbers: Trivial

$n = \overline A$ , therefore $A = A$ for all A between $0$ and $9$ .

Sum of $1$ digit solutions = sum of all the digits from $0$ to 9

$\frac{n(n+1)}{2} \Longrightarrow \frac{9(9+1)}{2} = 45$

For $2$ digit numbers:

We need to find numbers $\overline{AB}$ such that $(A+B) = AB$ .

$(A+B) = AB \Longrightarrow AB-A-B = 0$

$AB-A-B = A(B-1)-B = A(B-1)-B+1-1 = (A-1)(B-1)-1$

$(A-1)(B-1) - 1 = 0 \Longrightarrow (A-1)(B-1) = 1$

The only natural numbers that satisfy this equation are $A=B=2$ , giving us our one and only solution of $22$ .

Sum of $2$ digit solutions = $22$

Current sum: $45+22 = 67$

For $3$ digit numbers:

We need to find numbers $\overline{ABC}$ such that $(A+B+C) = ABC$ .

Everyone knows one example: $1+2+3 = 1\times2\times3 = 6$ . Since the order of the digits in our number doesn't matter, any permutation of $1$ , $2$ , and $3$ will suffice, giving us $123$ , $132$ , $213$ , $231$ , $312$ , and $321$ as solutions.

Are there any more solutions? Through a similar process as used above in the 2 digit problem, it can be shown that all solutions must be of the form: $(AC-1)(BC-1) = C^2+1$

We already know that this equation has solutions for $C=\{1,2,3\}$ , but brute force shows that plugging in the values $4$ through $9$ yields non-integer solutions, meaning that our above solutions are the only ones.

Sum of $3$ digit solutions = $1332$

Current sum: $67 + 1332 = 1399$

For $4$ digit numbers:

At this point it becomes easier to start simply to check what works and what doesn't. For $4$ digit numbers, it can be shown that no number can contain a 5 or higher and have solutions. To begin, we'll let $D > 4$ . If we let $A=B=C=1$ , then $ABCD = D$ and $A+B+C+D = D+3$ , meaning $ABCD < A+B+C+D$ . Now, let's allow $A > 1$ , $B > 1$ , and/or $C > 1$ . It's obvious that $ABCD$ will grow faster than $A+B+C+D$ as each variable increases, and plugging in our smallest possible increase for $C$ and $D$ , $2$ and $5$ , and letting $A=B=1$ yields $1\times1\times2\times5 = 10 > 9 = 1+1+2+5$ . Even the smallest increase in one variable was enough to make $ABCD > A+B+C+D$ . Since $A$ , $B$ , $C$ , & $D$ are all interchangeable, this is enough to prove that no digit can be greater than $4$ .

A little searching around gives us $1\times1\times2\times4 = 1+1+2+4 = 8$ as a solution. There are $4$ digits here with 2 digits that are the same so there are 4!/2! permutations, with solutions such as $1124$ , $1241$ , $4112$ , etc. A careful search shows that no solutions involving $3$ s exists so these are our only solutions.

Sum of $4$ digit solutions = $26664$

Final sum: $1399 + 26664 = \boxed{28093}$