# Sum of digits algebra problem

How would one go about finding all 3-digit positive integers $\overline{abc}$ with the property $\overline{abc}=abc(a+b+c)$, where $\overline{abc}$ would be the decimal representation of a number.

I have tried reducing to algebra, where one gets $100a+10b+c=abc(a+b+c)$, but i am at a complete loss as to what to do next!

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Brute-forcing netted me $135$ and $144$, but I'd sure like to see a better solution. –  Ｊ. Ｍ. Sep 5 '11 at 17:45

HINT $\$ Try casting out $9$'s and $11$'s to further constrain the solution space:
$$\rm a\:b\:c\:(a+b+c)\ \equiv\ a+b+c\ \ (mod\:\ 9),\ \equiv\ a-b+c\ \ (mod\:\ 11)$$
Since there are only 900 possibilities, this is trivial on a computer, but I suppose you want to do it by hand. Well, start with this. There obviously can't be any 0's. If the number is odd (i.e. $c$ is odd), then $a$ and $b$ are also odd. If $a$ or $b$ is 5, then $c=5$. Since $c$ divides $\overline{a+b+c}$, it must divide $100 a + 10 b = 10 (10 a + b)$. In particular, you can show that $c$ can't be 9, because if the sum of two digits is 9 one of them must be odd and the other even.