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Level Six > Number and Algebra

# What is s and t?

Achievement Objectives:

Specific Learning Outcomes:

Use guessing to make conjectures

Solve a problem using algebraic expressions

Description of mathematics:

This problem is the sixth of a series of six problems that develop from a specific stamp problem to a quite general one. The other problems in this series are 3c and 5c Stamps, Number, Level 3 4c and 7c Stamps, Number, Level 5, 5c and 9c Stamps, Number, Level 5, What is s?, Algebra, Level 6, and What is t?, Algebra, Level 6. The Level 4 and 5 problems look at specific problems following on the theme of the current problem. While we hint at generalisations in the Levels 4 and 5 problems we don’t follow these through until Level 6.

It is a good idea to make sure that the students have done earlier problems in this series before tackling this one. It will be especially valuable for them to have tried What is s?, Algebra, Level 6, before doing this question. It should provide them with ideas to try in the current problem.

Here we look at the earlier problem in this series from a different standpoint. Instead of giving two stamp values and finding a value from which all amounts of postage can be made, we give one stamp denomination and the value from which all amounts of postage can be made, and ask the students to find the other denomination. This is another example of an inverse problem. These occur all over the place and it is a useful skill to be able to solve problems from all angles.

In this problem, a certain amount of guessing is needed to solve the problem. Guided guessing is an important aspect of mathematics. In fact it is an aspect that is usually highly under-rated.

The interesting thing to note about this problem is that there is more than one answer. Sometimes that happens and students should learn not to be upset by it.

Going on to the Extension problem involves more guessing and also the constructing of proofs. Here this construction is not easy and may require some assistance. But it is a useful stage to go through as it will develop mathematical skills of conjecturing (guessing), proving and using algebra.

Required Resource Materials:
Copymaster of the problem (English)
Copymaster of the problem (Māori)
Activity:

### The Problem

The Otehaihai Post Office has two denominations of stamps. It can make up any amount of postage from 54c onwards. It can’t make up 53c. What are the values of the denominations?

### Teaching sequence

1. Pose the problem to the class. Discuss how they might solve the problem. Make sure that they understand that there are two key ideas here. The values of s and t have to be such that they produce 54, 55, 56, …, but that they don’t 53.
2. Let the students work on the problem with a partner.
3. Then if the students are stuck you might suggest that they experiment with various values of s and t, the values of the two denominations (and keep a record of their work as it will be useful later).
4. Share students’ solutions with the whole class. Note any different answers and any different methods of solution.
5. It is worth letting as many students as possible attempt the Extension.
6. Get the students to write up what they have discovered.

#### Extension to the problem

Make up a problem of your own like the one above using values other than 54 and 53.

Is it possible to make up a problem where 54 is replaced by an odd number?

### Solution

The method of proof here is essentially that of What is s?, Algebra, Level 6.

Again this can be done by guessing and checking. It should produce four answers (see below).

 s 2 3 4 7 t 55 28 19 10

A nicer method than the one above, is to know (or guess) that in general the lowest ‘point’ is (s – 1)(t – 1 ). However, this is quite difficult to prove. So we have not required that here. But you might like to know that the result was first proved by an English mathematician of the 19th Century called Sylvester. This can then be equated to 54. Now to solve this equation, note that the factors of 54 are 1, 2, 3, 6, 9, 18, 27, 54. So, in pairs, s – 1 and t – 1 have to equal 1 and 54, 2 and 27, 3 and 18 and 6 and 9. (Now check that you can’t get 53 in any of these cases.)

It is worth noting that (s – 1)(t – 1) can never be odd. This could only happen if s – 1 and t – 1 were both odd. That would mean that s and t were both even. But then they would have a common factor of 2. However, a linear combination of even numbers can never be odd. So it would not be possible to get all values of postage from some point on.

AttachmentSize
WhatisSandT.pdf36.53 KB
WhatisSandTMaori.pdf41.25 KB

## What is t?

Use guessing to make conjectures

Solve a problem using algebraic expressions

## What is s?

Use guessing to make conjectures

Solve a problem using algebraic expressions

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