How many $3$ digit numbers with digits $a$,$b$ and $c$ have $a=b+c$ My question is simple to state but (seemingly) hard to answer. How many $3$ digit numbers exist such that $1$ digit is the sum of the other $2$. I have no idea how to calculate this number, but I hope there is a simple way to calculate it. Thank you in advance.
EDIT: The first digit should not be $0$

Assuming a digit is an element of $\{0,1,2,3,4,5,6,7,8,9,10\}$ we have three cases for $a,b,c$ to see:


*

*$a=b=c=0$. All easy here, yields $1$ combination.

*$b=c\ne 0$. $a=2b$, so $b<5$ giving us $4$ choices (digits $1$ to $4$). The position of $a$ uniquely determines the code, so multiply b $3$ to get $4\cdot 3 = 12$ combinations

*$b\ne c$. We assume $a\ge b>c$ and chose $c$ first. Since $b+c < 10$ and $c<b$ $a \ge 2c$ so $c\le 4$.
$$\begin{align*}
c=4 & \Rightarrow b=5, a=9 & 1\\
c=3 & \Rightarrow b\in\{4,5,6\} & 3\\
c=2 & \Rightarrow b\in\{3,4,5,6,7\} & 5\\
c=1 & \Rightarrow b\in\{2,\ldots, 8\} & 7\\
c=0 & \Rightarrow b\in\{1,\ldots, 9\} & 9 (\text{only $2$ distinct digits here})
\end{align*}$$
totaling $16+9$ combinations, times $3! = 6$ for all but the $9$ we get $16\cdot 6 + 9 \cdot 3 = 123$


Summing up we have $1+12+123 = 136$ possibilities.
