Exercise (1.4).6

Assume there are one hundred pence in the pound (£). Using just 5p (£0.05) and 10p (£0.10) pieces, how many of each do you need in order to pay a parking meter charge of £3.10.

We encode the problem as followed with $f$ being the number of 5p coins and $t$ the 10p coins

$$ 5f + 10t = 310 $$

The $gcd(5,10)=5$ divides 310, so the equation has integer solutions.

In fact, by inspection we can see 310 is a multiple of 10, so 31 10p coins is a solution. 

Let's find other solutions. We already have one solution $f_0 = 0, t_0 = 31$. The general solution is

$$ f = 0 + \frac{10}{5}n = 2n, \quad t = 31 - \frac{5}{5}n = 31-n$$

for any integer $n$.

So need $f$ and $t$ to be non-negative, which gives us the following inequalities

$$ 2n \ge 0 $$

$$ 31 - n \ge 0$$

That is

$$ 0 \leq n \leq 31 $$

This creates 32 different solutions.

n	f=2n	t=31-n	5f+10t
0	0	31	310
1	2	30	310
2	4	29	310
3	6	28	310
4	8	27	310
5	10	26	310
6	12	25	310
7	14	24	310
8	16	23	310
9	18	22	310
10	20	21	310
11	22	20	310
12	24	19	310
13	26	18	310
14	28	17	310
15	30	16	310
16	32	15	310
17	34	14	310
18	36	13	310
19	38	12	310
20	40	11	310
21	42	10	310
22	44	9	310
23	46	8	310
24	48	7	310
25	50	6	310
26	52	5	310
27	54	4	310
28	56	3	310
29	58	2	310
30	60	1	310
31	62	0	310