A collection of bars costs £2 and a collection of rolls costs £3. List the number of bars and rolls you can purchase with £20, leaving no change. You must buy at least one of each.
We encode the scenario as follows, with $b>0$ being the number of bars, and $r>0$ the number of rolls.
$$ 2b + 3r = 20$$
The coefficients are small enough for us to find solutions by brute force, but we'll follow the procedure we'd use for larger coefficients.
Here $\gcd(2,3)=1$, which divides 20, so the equation has integer solutions.
By inspection a solution is $b_0 = 4, r_0 = 4$. A general solution is
$$ b = 4 + 3t, r = 4 - 2t $$
for any $t$.
However, since both $b$ and $r$ must be greater than zero we have the following inequalities
$$ 4 + 3t > 0 $$
$$ 4 - 2t > 0 $$
Re-arranging
$$ t > -\frac{4}{3} $$
$$ 2 > t $$
That is
$$ - \frac{4}{3} < t < 2$$
The only values of $t$ that lead to integer $b$ and $r$ are $t=-1, 0 and 1$. There are therefore 3 solutions:
- $t=-1$ gives $b=1, r = 6$, that is one bar and six rolls.
- $t = 0$ gives $b=4, r = 4$, that is four bars and four rolls.
- $t=1$ gives $b=7, r=2$, that is seven bars and two rolls.