Diophantus I

Achievement Objectives
NA6-5: Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns.
Student Activity

Mathematical puzzles have fascinated people throughout the ages.
These were often expressed in verse or as riddles like this:

When first the marriage knot was tied between my wife and me,
Her age did mine as far exceed as three plus three does three;
But when three years and half three years we man and wife had been
Our ages were in ratio then as twelve is to thirteen.

How old were they on their wedding day?

[Diophantus, a Greek mathematician]

 

Specific Learning Outcomes

Solve a problem in a number of ways, including using algebraic expressions

Description of Mathematics

Students may approach this problem using guess and check or by being systematic in some way. However, students should be supported to understand, and to use, an algebraic method of solution, recognising its efficiency.

The problem was apparently engraved on a tombstone in the time of the Greek mathematician Diophantus who lived in Alexandria somewhere between 150 BC and 364 AD. Diophantus wrote a thirteen-volume set of books called Arithmetica of which only six  have survived. He was interested in problems that had whole number solutions. Accordingly, equations of this type are called Diophantine equations.

Other related problems include:  A Lady’s Age and Diophantus II Algebra Level 6.

Required Resource Materials
Activity

The Problem

Mathematical puzzles have fascinated people throughout the ages. These were often expressed in verse or as riddles like this:

When first the marriage knot was tied between my wife and me,
Her age did mine as far exceed as three plus three does three;
But when three years and half three years we man and wife had been
Our ages were in ratio then as twelve is to thirteen.

How old were they on their wedding day?

[Reported to have been inscribed on the grave of Diophantus, a Greek mathematician from Alexandria (100BC approx)]

Teaching sequence

  1. Have students consider the historical nature of the puzzle with questions such as:
    Who is the most famous person you know who was born over 100 years ago?… over 1000 years ago? … over 2000 years ago?
    Who was the dominant power in Europe 2000 years ago?
    Where was the European centre of learning then?
  2. Read Diophantus’ poem and, with the students, tease out it's meaning. 
  3. Establish understanding of mathematic references with questions such as:
    What does ‘exceed’ mean?
    What does 'the ratio as twelve is to thirteen' mean?
  4. Have students work on the problem, recording their working as they do so. Monitor their progress.
  5. When a group finds a solution using an algebraic approach, encourage them to work on the Extension problem.
  6. Allow time for several groups to share their solutions and have the class discuss these together.
  7. Make time for students to write up two ways of solving the problem. 

Extension

Write a problem about your own age, or someone else’s age, for members of the class to solve.

Using online sources, find out as much as you can about Diophantus or other ancient Greek mathematicians.

Solution

Three possible approaches to this problem are given here.

Method 1. Guess and Improve. Try an initial guess. If this doesn’t work, then try another. Use the first guess to make the second one a better guess.

Suppose he was 30, then his wife was 33. After 3 + ³/₂ = 4 ¹/₂ years, he was 34 ¹/₂, and his wife was 37 ¹/₂.
Now the ratio of his age to hers is 34 ¹/₂: 37 ¹/₂, which equals 1:1.086 or 12:13.043. This is close but not close enough.

So try his age as 33. Then his wife’s age was 36. After 4 ¹/₂ years, he became 37 ¹/₂, and she became 40 ¹/₂.
The ratio now is 1:1.08 or 12:12.96. Since 12.96 is on the other side of 13 to 13.043, the next guess should be between 30 and 33.

So try 31.5 for Diophantus’age. His wife would then be 34.5. After 4 ¹/₂ years, he is 36, and she is 39. The ratio between their ages is now 36:39, which equals 12:13. This is correct.

Method 2. Test every possible combination. The most efficient way to do this is using a computer program.

Method 3. Use algebra. Let his age be D. Then his wife’s age is D + 3.  An equation can be set up accordingly:

(D + 4 ¹/₂)/[(D + 3) + 4 ¹/₂] = 12/13.

Rearranging gives 13(D + 4 ¹/₂) = 12[(D + 3) + 4 ¹/₂].
'Tidying up' we have 13D + 58.5 = 12D + 90.
So D = 31.5,
as we found by Method 1.

Hence, when they got married, Diophantus was 31.5 and his wife 34.5.

This method is more efficient than Methods 1 and 2. Support students to see the advantage of using algebra.

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