Determine All positive solutions: $54x+21y=906$ Determine All positive solutions: 
$54x+21y=906$ 
so first I tried to see if the gcd$(54,21)|906$ which is true since the gcd of $(54,21)=3$
Then I tried to find one solution by writing a linear equation of $a$ and $b$
So far I had 
$3=12-9(1)$
$3=12-[21-12(1)](1)$
$3=2(12)-1(20)$
$3=2(54-21(2))-1(20)$
$3=2(54)+...$
Now this last line is a problem. Why do I have an extra $20$? Did  I do something wrong? 
 A: I guess that the question is about positive integral solutions.
This problem is known - in general - as the problem of determining the so-called "Numerical semigroup" generated by two positive integers $a, b$. See, this math overflow post https://mathoverflow.net/questions/121826/on-the-set-of-numbers-generated-by-integer-linear-combination-of-two-real-number, and the wikipedia page http://en.wikipedia.org/wiki/Numerical_semigroup.
When $a$ and $b$ are coprime, one can express every positive integer $s$ that is larger or equal to $(a-1)(b-1)$ as solution $s= xa+yb$ with $x, y\in \mathbb{N}$. From this you can easily get a statement for the case when they are not coprime.
Note: I think that this problem is also known as "coin problem".
Here you can find an explicit guide for finding the positive solutions:
https://www.cs.cmu.edu/~adamchik/21-127/lectures/divisibility_5_print.pdf
A: First you need to find a solution. As you note, the coefficients have a GCD of 3, so let's divide that out, getting $18x+7y=302$.  For small numbers, you can just do it by hand.  We $x \le 16$ because $17 \cdot 18=306$ is too large.  We note that $16 \cdot 18 + 2 \cdot 7=302$  Now for smaller solutions, we can trade $7$ in $x$ for $16$ in $y$.  Since we can only do the trade twice, there are three solutions $(16,2),(9,18),(2,34)$ 
