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 $0$
 A: One digit ($a$) will be at least as large or larger than the other two digits ($b, c$).   For every combination of $a$ and $b$ there can be only one value of $c$.
Another restaint is that the first digit cannot be 0, so we need consider the placement of the largest digit.
If the first digit is the largest, then we must count all pairs of of $a\in\{1,\ldots,9\}$ and $b\in\{0,\ldots, a\}$.
If the largest digit is the second or third digit then, as we have already counted all the cases when it is equal to another digit, we must count all pairs of $b\in\{1,\ldots,8\}$ and $a\in\{b+1,\ldots, 9\}$, twice (for the two placements).
The total count of is then: $$\begin{align}
 & =\sum\limits_{a=1}^9 \sum\limits_{b=0}^a 1 + 2\times \sum\limits_{b=1}^8 \sum\limits_{a-b=1}^{9-b} 1
\\ & = \sum\limits_{a=1}^9 (a+1) + 2\times \sum\limits_{b=1}^8 (9-b)
\\ & = \tfrac{9(9+1)}{2} + 9 + 2(9\cdot 8 -\tfrac{8(8+1)}{2})
\\ & = 126
\end{align}$$
A: Various cases are:
$$\begin{array}{|c|c|}\hline
(a,b,c)|a+b=c&\text{total numbers}\\\hline
(1,1,2),(2,2,4),...(4,4,8)&3!/2!\times4\\\hline
(1,2,3)\text{ to }(1,8,9)&3!\times7\\
(2,3,5)\text{ to }(2,7,9)&3!\times5\\
(3,4,7)\text{ to }(3,6,9)&3!\times3\\ 
(4,5,9)&3!\times1\\\hline
(1,0,1)\text{ to }(9,0,9)&2\times9\\\hline
\end{array}$$
Total ways: $$3!/2!\times4+3!\times(1+3+5+7)+2\times9=12+6\times4^2+18=126$$
A: 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.
A: For some, this might not be the most intuitive way, but it is a way to solve this problem.
If $a=b+c$, then if a is an even number it gives us 2 exceptions, and if a is odd it gives 1 exception. Will keep exceptions in a seperate bucket.
The maximum sum that can be reached is 9. 
So if $(b,c,a)$ represent the set of possible number then, 
$(1,8,9),(2,7,9),(3,6,9),(4,5,9)$ and then $b,c$ change place. That's $4\times 2 =8$ numbers. And since places are not fixed, therefore its $8\times3=24$ numbers + 1 exception $(909)$
Now, For 8  $(1,7,8),(2,6,8),(3,5,8)$ that's $3\times2\times3=18$ numbers +2 exceptions $(448,808)$
For 7  $(1,6,7),(2,5,7),(3,4,7)$ that's $3\times2\times3=18$ numbers +1 exception $(707)$
For 6  $(1,5,6),(2,4,6)$ that's $2\times2\times3=12$ numbers +2 exception $(606,336)$
For 5  $(1,4,5),(2,3,5)$ that's $2\times2\times3=12$ numbers +1 exception $(505)$
For 4  $(1,3,4)$ that's $2\times1\times3=6$ numbers +2 exception $(404,224)$
For 3  $(1,2,3)$ that's $2\times1\times3=6$ numbers +1 exception $(303)$
So far, 96 numbers. Now, lets look at the combinations of number put in exception and will count and add seperately.
$(909,990,448,844,484,808,880,707,770,606,660,336,363,633,112,121,211,202,220,101,110)$
Total count of number :$117$
A: 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.
