Finding the values of 9 variables given 4 expressions and a constraint on the values of the variables Let's say I have $9$ variables $(a, b, c, d, e, f, g, h, i)$ and I know they are all different and they all have values between $1$ and $9$ included. Basically these variables will each have a different value between 1 and 9.
So:
$$a, b, c, d, e, f, g, h, i \in [1, 9]$$
$$a \neq b \neq c \neq d \neq e \neq f \neq g \neq h \neq i$$
I also know that:
$$a + b + c + d + e = 22$$
$$a + b + f + g + h = 22$$
$$d + e + g + h + i = 22$$
How do I determine the values of $a, b, c, d, e, f, g, h, i$?
 A: The first thing to notice is that there is likely to be a lot of solutions since given any solution we can interchange various pairs of letters and keep the same sums.
The letters $c,f,i$ are special in that they all occur in only one equation and can all be interchanged with each other.
The sums of five numbers adding to $22$ and containing a $9$ are
$$1,2,3,7,9$$
$$1,2,4,6,9$$
$$1,3,4,5,9$$
Any one of these  solutions has three elements in common with any other. This is not the case for your equations and so the $9$ can only be in one of the equations i.e. $9$ must be one of $c,f,i$.
Similarly, $8$ must also be one of $c,f,i$ and then $7$ also is one of $c,f,i$. Without loss of generality we can suppose $$c=9,f=8,i=7. $$
The equations are now easy to solve. For example:
$abcdefghi$ are respectively $249168357$.
Number of solutions
There are $3!$ possibilities for $c,f,i$. For each of these possibilities $a+b,d+e,g+h$ are then determined as $6,7,8$ in a particular order. These can be split as either $1+5,3+4,2+6$ or $2+4,1+6,3+5$. The total number of solutions is therefore $$3!\times2\times2^3=96.$$
A: Adding all the equations gives $$2a+2b+c+2d+2e+f+2g+2h+i=66$$
Since twice the sum of all the letters is  $90$ we have $c+f+i=24$ and therefore $c,f,i$ are $7,8,9$ in some order.
Now use either of the earlier answers!
