Vladimir chose three digits $a, b$ and $c$ such that $a>b>c>0$ and with them formed the numbers $abc$, $cba$ and $cab$. Note that $abc$ is not the product of $a, b$ and $c$, but rather the number formed by the digits $a, b$ and $c$. For example, if $a=1, b=2$ and $c=3$, $abc$ will be the number 123.

After choosing these three digits $a, b$ and $c$, Vladimir noticed that one of the numbers formed was equal to the sum of the other two. Find the numbers formed by Vladimir.
Solution

Since $a>c$, the number $abc$ is greater than the numbers $cba$ and $cab$, and thus we must have $abc=cba+cab$. Therefore,

$$
100 \cdot a + 10 \cdot b + c = (100 \cdot c + 10 \cdot a + b) + (100 \cdot c + 10 \cdot b + a)
$$

and thus $89a = 199c + b$.

Note that $10 > a = (199 \cdot c + b) / 89 > 2c$, and that $a = (199 \cdot c + b) / 89c$. We will consider cases based on the value of $c$.

- If $c=1$: We should have $2 < a < 3$, and there is no digit $a$ that satisfies this.
- If $c=2$: We should have $4 < a < 6$. Then $a=5$ and $b=89 \cdot a - 199 \cdot c = 47$. There is no solution in this case.
- If $c=3$: We should have $6 < a < 9$. Then $a=7$ or $a=8$, and $b=89 \cdot a - 199 \cdot c \geq 89(7) - 199(3) = 26$. There is no solution in this case.
- If $c=4$: We should have $8 < a < 12$. Then $a=9$ and $b=89 \cdot a - 199 \cdot c = 5$. In this case, the numbers formed by Vladimir are 954, 459, and 495.