Find all the three digit numbers $\overline{a b c}$ such that

$$
\overline{a b c}=a b c(a+b+c)
$$
We will show that the only solutions are 135 and 144.

We have \(a > 0, b > 0, c > 0\) and

\[
9(11a + b) = (a + b + c)(abc - 1)
\]

- If \(a + b + c \equiv 0 \pmod{3}\) and \(abc - 1 \equiv 0 \pmod{3}\), then \(a \equiv b \equiv c \equiv 1 \pmod{3}\) and \(11a + b \equiv 0 \pmod{3}\). It follows now that

\[
a + b + c \equiv 0 \pmod{9} \text{ or } abc - 1 \equiv 0 \pmod{9}
\]

- If \(abc - 1 \equiv 0 \pmod{9}\)

we have \(11a + b = (a + b + c)k\), where \(k\) is an integer

and it is easy to see that we must have \(19\).

Now we will deal with the case when \(a + b + c \equiv 0 \pmod{9}\) or \(a + b + c = 9l\), where \(l\) is an integer.

- If \(l \geq 2\) we have \(a + b + c \geq 18, \max\{a, b, c\} \geq 6\) and it is easy to see that \(abc \geq 72\) and \(abc(a + b + c) > 1000\), so the case \(l \geq 2\) is impossible.
- If \(l = 1\) we have

\[
11a + b = abc - 1 \text{ or } 11a + b + 1 = abc \leq \left(\frac{a + b + c}{3}\right)^3 = 27
\]

So we have only two cases: \(a = 1\) or \(a = 2\).

- If \(a = 1\), we have \(b + c = 8\) and \(11 + b = bc - 1\) or \(b + (c - 1) = 7\) and \(b(c - 1) = 12\) and the solutions are \((a, b, c) = (1, 3, 5)\) and \((a, b, c) = (1, 4, 4)\), and the answer is 135 and 144.
- If \(a = 2\) we have \(b(2c - 1) = 23\) and there is no solution for the problem.