Diophantine equations (named for Diophantus of Alexandria, 3rd century Greek mathematician known sometimes as the father of algebra), which are linear equations with integer solutions, provide an excellent extension path for secondary students. The following illustrative problem is based on sport.
Example 1.1
An AFL team has 44 points and has had 14 scoring shots. How are these made up?
Note
Of course we need to know the scoring system for Australian Football. Basically a team scores 6 points for a goal, where the ball goes between the central posts of the boot without being touched, while 1 point for a behind, where the ball otherwise goes through the two outer of the four posts.
Solution 1.1 Alternative 1
The easiest way is to experiment around the likely numbers, as the solution will be unique. If 5 goals this has 30 points so 14 points would be needed, too many. Upping to 6 goals this gives 36 points, requiring 8 behinds, and this is the solution.
Solution 1.2 Alternative 2
More formally, if G is the number of goals and B is the number of behinds, then
6G+B=44 ...(1),
because the total point score is 44, and
G+B=14 ...(2),
because the total number of shots is 14. Subtracting (2) from (1) gives 5G=30, i.e. G=6, leaving, from (2), B=14 - 6 = 8, as in Alternative 1.
Example 1.2
Red rose plants are for sale at $3 each and yellow ones for $5 each. A gardener wants to buy a mixture of both types (at least one of each) and decides to buy 13 in total, with more yellow ones than red ones. The number of dollars he spent could be
(A) 51 (B) 67 (C) 65 (D) 58 (E) 57.
Strategy
Because of the finite nature of the problem, the student could canvass all the possibilities, working out the amounts when the number of yellow flowers varies from 7 to 12 inclusive (yielding all odd numbers between 53 and 63 inclusive). However, the problem could be done algebraically. Here, the students are challenged to define suitable variables and construct the necessary functions. Logical thinking, along with strict attention to the conditions of the problem, will lead students of average ability to the solution.
Solution 1.2 Alternative 1
Suppose the gardener buys x red rose plants, and hence 13-x yellow ones (1≤ x≤ 6 since there is at least one of each type and there are more yellow ones than red ones). Then the total amount of money, in dollars, which is spent, is 3x+5(13-x)=65-2x (remembering 1≤ x≤ 6).
The possible numbers of dollars are thus 63, 61, 59, … , 55, 53. The only answer of these on the offered list is 57, hence the answer is (E).
Solution 1.2 Alternative 2
Or we can directly list the 6 obvious allowed cases and calculate the cost of each.
12 yellow, 1 red, cost = 12 x 5 + 1 x 3 = 63,
11 yellow, 2 red, cost = 11 x 5 + 2 x 3 = 61,
10 yellow, 3 red, cost = 10 x 5 + 3 x 3 = 59,
9 yellow, 4 red, cost = 9 x 5 + 4 x 3 = 57,
8 yellow, 5 red, cost = 8 x 5 + 5 x 3 = 55,
7 yellow, 6 red, cost = 7 x 5 + 6 x 3 = 53.
The only answer of these on the offered list is 57, hence the answer is (E).