
NT4. Find all the three digit numbers $\overline{a b c}$ such that

$$
\overline{a b c}=a b c(a+b+c)
$$


## Solution

We will show that the only solutions are 135 and 144 .

We have $a>0, b>0, c>0$ and

$$
9(11 a+b)=(a+b+c)(a b c-1)
$$

- If $a+b+c \equiv 0(\bmod 3)$ and $a b c-1 \equiv 0(\bmod 3)$, then $a \equiv b \equiv c \equiv 1(\bmod 3)$ and $11 a+b \equiv 0(\bmod 3)$. It follows now that

$$
a+b+c \equiv 0(\bmod 9) ; \text { or } a b c-1 \equiv 0(\bmod 9)
$$

- If . $a b c-1 \equiv 0(\bmod 9)$

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

and is easy to see that we must have $19$.

Now we will deal with the case when $a+b+c \equiv 0(\bmod 9)$ or $a+b+c=9 l$, 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 $a b c \geq 72$ and $a b c(a+b+c)>1000$,so the case $l \geq 2$ is impossible.
- If $l=1$ we have

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

So we have only two cases: $a=1$ or $a=2$.

- If $a=1$, we have $b+c=8$ and $11+b=b c-1$ 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 $b(2 c-1)=23$ and there is no solution for the problem.
