Diophantus I

The Problem

Mathematical curiosities and puzzles have fascinated people throughout the ages. These were often expressed in verse or as riddles. This one is supposed to have come from the grave of an ancient Greek mathematician who lived in Alexandria.

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 (100BC approx)]

Teaching sequence

1. Talk about historical things. (If you can find some relevant pictures to show the class, so much the better.)
   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. Recite Diophantus’ poem. Make sure they have some idea of what it is about.
   Do you know what ‘exceed’ means?
   What about the ratio twelve is to thirteen?
3. Get the class to work on the problem in groups of two or four.
4. Circulate around the class and check on progress. If a group has finished using an algebraic approach, then let them try the Extension problem.
5. Allow time for several groups to report on their answer and the ways that they solved the problem. If there is not time to look at the Extension problem ask them to take it home and get their parents’ help.
6. At some stage let the class write up two ways of solving the problem in their mathematics book.

Extension to the problem

Can you make up a problem about your own age or about someone else’s? Give it to another member of the class to solve.

Use the Internet to find out as much as you can about Diophantus or some of the other ancient Greek mathematicians.

Solution

As in a number of other problems, we give three possible approaches to this problem. All of your students should be able to use Method 1. The more sophisticated method is Method 3 where we use algebra. Hopefully most of your class can at least get started on the algebraic approach.

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 1/2which 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 1:1.08or 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. BINGO!

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

Method 3. Use algebra. Let his age be D (more suggestive than x). Then his wife’s age is D + 3. So we can set up an equation. What we get is

(D + 41/2)/[(D + 3) + 41/2] = 12/13.

Rearranging gives 13(D + 41/2) = 12[(D + 3) + 41/2].
Tidying up a bit 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 a lot more efficient than the other two. This is often the case with algebra. It’s really quite a powerful technique.