Find number of positive integer solutions $(x,y,z)$ for the following equation:
$19x + 11y + 8z = 240$
I divided the equation by $8$ and then tried to equate remainders. It yields that $3(x + y) = 8k$ for some constant $k$ or $x + y$ is a multiple of 8. Can't choose which combinations of $x,y$ will do the work.
Solved!
For $ x + y = 8 $ All ordered pairs are allowed. Corresponding values of $z$ are within domain of positive integers. This gives 7 solutions.
For $ x + y = 16 $ Only 7 pairs of $ x,y $ allowed. They are $ (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9) $ Pairs $(8,8)$ and beyond yield negative values for $z$. Therefore total of 14 solutions.
I had a very similar method, which takes advantage of $19-11=8$ and also the factor $8$ generally. Rewrite as $$11(x+y)+8(x+z)=240=8\times 30$$
$x+y$ must be divisible by $8$ and the only possibilities are $x+y=8, 16$. In the first of these cases $x+y=8$ has seven solutions. In the second we find $x+z=8 (=30-2\times 11)$ which has seven solutions. All these solutions evidently give a positive integer value for the missing number, so there are fourteen solutions.