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

Step 1: Since $a,b,c$ are digits, we must have $0\le a,b,c\le 9$, with the leading digit restriction $1\le a\le 9$.

Step 2: The relation is $c=a+b$. Since $c$ is a digit, we must have:

Step 3: For each chosen value of $a\in \{1,2,\dots ,9\}$, the digit $b$ can take any integer value from $0$ up to $9-a$. This gives exactly $10-a$ choices for $b$. Once $a$ and $b$ are chosen, $c$ is uniquely determined as $a+b$.

Step 4: Summing the choices over all valid values of $a$:

So there are 45 such 3-digit numbers.