How many 3-digit numbers $$\overline{abc}$$ in base $$10$$ are there with $$a \neq 0$$ and $$c = a + b$$?

Let the three-digit number be $$\overline{abc}$$, where $$a,b,c$$ are its decimal digits.

Digit conditions:
1. $$a \neq 0$$ because the number is three-digit, so $$a \in \{1,2,\dots ,9\}$$.
2. $$b,c \in \{0,1,\dots ,9\}$$ (any decimal digit).
3. Given condition: $$c = a + b$$.

Since $$c$$ itself must be a single digit, we need $$a + b \le 9$$. Thus the problem reduces to counting ordered pairs $$(a,b)$$ satisfying

$$a \in \{1,2,\dots ,9\}, \qquad b \in \{0,1,\dots ,9\}, \qquad a + b \le 9$$.

Fix a particular value of $$a$$ and determine how many $$b$$ are possible.

Continuing similarly, we get the following counts:

$$ \begin{aligned} a = 1 &\Rightarrow 9 \text{ choices for } b\\ a = 2 &\Rightarrow 8 \text{ choices for } b\\ a = 3 &\Rightarrow 7 \text{ choices for } b\\ &\ \vdots \\ a = 8 &\Rightarrow 2 \text{ choices for } b\\ a = 9 &\Rightarrow 1 \text{ choice for } b\\ \end{aligned} $$

The numbers of choices form an arithmetic progression $$9,8,7,\dots ,1$$.

Total number of valid pairs $$(a,b)$$ is therefore the sum of this progression:

$$ \text{Total} = 9 + 8 + 7 + \dots + 1 = \frac{9 \times 10}{2} = 45. $$

For every such pair $$(a,b)$$ the digit $$c$$ is fixed as $$c = a + b$$, so each pair gives exactly one valid three-digit number.

Hence the total count of three-digit numbers satisfying $$c = a + b$$ is $$\mathbf{45}$$.