<s>[INST] 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 $0$
 [/INST] My answer is similar to that of @shooting-squirrel. Edit: Now it gives the same answer, silly mistakes :D
Consider the number $\overline{abc}$, we have three cases:
Case 1: $a=b+c$, then $a$ can be $1,\dots,9$. Now, for each value of $a$, there are $a+1$ values of $b$: $0,1,\dots, a$. Each of which gives one number. Total numbers in this case is 
$$\sum^{9}_{a=1}(a+1)= \frac{(10+2)\cdot 9}{2}=54.$$
Case 2: $b=a+c$. Since $b\ge a$, $b$ can only be $1,\dots,9$. For each $b$, there are $b-1$ values of $a$: $1,\dots, b-1$. ($a=b$ was counted in Case 1).
Each of which gives one number. Total number in this case is
$$\sum^9_{b=1}(b-1)=\frac{8\cdot 9}{2}=36.$$
Case 3: $c=a+b$. This case is similar to Case 2, with $c$ plays the role of $b$. So we have $36$ numbers in Case 3.
Total: Case 1 + Case 2 + Case 3 = $54+36+36=\color{red}{\mathbf{126}}$ numbers.
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