This problem solving activity has a number and algebra (equations and expressions) focus.
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]
Students may approach this problem by using guess and check or by being systematic in some way. However, students should be supported to understand, use, and recognise the efficiency of an algebraic method of solution.
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 Level 6 Number and Algebra (Equations and Expressions) problems include: The Lady’s Age and Diophantus II.
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)]
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.
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.
Printed from https://nzmaths.co.nz/resource/diophantus-i at 3:20am on the 29th March 2024