The lady's age

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.

A lady being asked her age, gave the following reply:

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?

[A Lady’s Diary, 1788]

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. 
   You could ask:
   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? (The lady’s age.)
3. Have students work on the problem. Encourage them to record their solutions to enable them to share with others. 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 can be 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.