Adding digits

We are given

$\begin{cases} 2a + b = 10b + c & \qquad (1) \\ b + c = b & \qquad (2) \end{cases}$

It is easy to see from $(2)$ that $c = 0$ . Our first equation becomes $2a = 9b$ , and the only integer solutions for this occur when $a = 9n$ and $b = 2n$ for some natural number $n$ . $\text{lcm} (2, 9) = 18 \Longrightarrow a = 9, b = 2$ produces the only one-digit integer solutions.

After doing a quick check that this solution satisfies the given conditions, we conclude that $\overline{abc} = \boxed{920}$ .

Since the sum of bc (I don't know how to indicate that b and c are digits and not two numbers I'm multiplying) equals b, c must equal 0.

Now, the maximum value for the sum of 3 single digit positive numbers is 27. So, b is either equal to 1 or 2. If b is equal to 1, then the sum of the digits of aba is 10. But the sum of the digits of aba is b + 2a, which equals 1 + 2a, which must be an odd number. Thus, this sum cannot be 10.

So, b must be 2, and the sum of aba must be 20. Quickly, we see that a = 9.

So, the number abc is 920.