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 curiosities and puzzles have fascinated people throughout the ages. These were often expressed in verse or as riddles.
Here is one of these.

If to my age there added be;
One half, one third and three times three
Six score and ten the sum you’d see

Now pray tell what my age may be?

Specific Learning Outcomes

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

Description of Mathematics

This problem is based on a 1788 riddle. To solve it, students may guess and check, guess and improve, or test every combination. However, students should be encouraged and supported to see that using algebra is the most efficient strategy.

Two related problems which demonstrate the power of algebra are: Level 6, are Diophantus I and Diophantus II

For further reading, see:  A History of Mathematics by Carl Boyer.
H.E. Dudeney’s Amusements in Mathematics, Nelson, 1919,
W.W.R. Ball’s Mathematical Recreations and Essays, Macmillan, 1939
E. Lucas (four volumes of his Récréations Mathématiques were published by Gautier-Villars, between 1883 and 1894),

Required Resource Materials
Activity

### The Problem

Mathematical curiosities and puzzles have fascinated people throughout the ages. These were often expressed in verse or as riddles. Here is one of these.

If to my age there added be;
One half, one third and three times three
Six score and ten the sum you’d see
Now pray tell what my age may be?

### Teaching sequence

1. Discuss with students aspects of history that pertain to the problem. In particular, highlight that a woman's modesty about her age was commonplace.
Who is the most famous person you know who was born over 50 years ago?
When was the Eighteenth Century?
Can you tell us of something that happened in the Eighteenth Century?
2. Pose the riddle from the Lady’s Diary. Ensure it is understood. Ask:
Why do you think the problem was invented?
What is a ‘score’? What is ‘six score and ten’?
What does ‘one half’ and ‘one third’ refer to?
3. Have students work on the problem, and encourage them to record their solutions to enable them to share with others.
If a group finishes using an algebraic approach, make the Extension problem available.
4. Allow time for several groups to share their solutions. Discuss, and highlight the efficiency of the algebraic approach.

### Extension

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

Find online (books that contain) mathematical problems, puzzles or recreations. Seek ideas beyond Western literature, such as those in Chinese and Japanese writing.

### Solution

The approaches used by students will vary.
Three possible approaches are given for this problem.

Method 1. Guess and Improve.
Because Guess and Check is very inefficient,  Guess and Improve is used to hone the answer.

Six score and ten is 130, so we have to guess an age for the lady so that it plus one half of it plus one third of it plus three times three equals 130.

Suppose the lady was 48. Then one half of 48 is 24 and one third of 48 is 16. Clearly three times three is 9. So we have 48 + 24 + 16 + 9 = 97. This is too small. So try a bigger number.

Suppose she was 90. Then 90 + 45 + 30 + 9 = 174 - too big. So we have to try something bigger than 48 and smaller than 90.

Suppose she was 60. Then 60 + 30 + 20 + 9 = 119 - too small. So we want a number between 60 and 90.

Suppose she was 66. Then 66 + 33 + 22 + 9 = 130: AHA!!

Method 2. Test every possible combination. This is a rather laborious way of doing the problem. However, a computer programme might run through all numbers from 1 upwards, checking the criteria of the problem.

Method 3. Use algebra. Let the lady's age be a.

 Then we have to solve So Hence This gives or Tidying up we get a + a/2 + a/3 + 9 =  a + a/2 + a/3 =  (6a + 3a + 2a)/6 =  11a/6 = a = a = 130. 121. 121. 121 (121 x 6)/11. 66

This method is a lot more efficient than the other two. This is often the case with algebra.

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