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Level Six > Number and Algebra

# What is t?

Achievement Objectives:

Specific Learning Outcomes:

Use guessing to make conjectures

Solve a problem using algebraic expressions

Description of mathematics:

This problem is the fifth 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 s and 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. Once they have mastered that problem they will be able to use it as a model for this one.

Here we look at the earlier problem in this series from a different standpoint. This time we don’t give two stamp values and then try to find the value from which all amounts of postage onwards can be made. Instead we give one stamp denomination and the smallest postal value from which all amounts onwards 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. 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 has 4c stamps and stamps of some other value. It can make up any amount of postage from 42c onwards. It can’t make up 41c. What is the value of the other denomination?

### 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 value of t has to be such that it produces 42, 43, 44 and 45, but that it doesn’t produce 41.
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 t (and keep a record of their work as it will be useful later).
4. Share students’ solutions with the whole class. Note any different methods of solution.
5. In the Extension, the students will probably need the following hints: (i) experiment with different values of t to get some ideas; and (ii) in the proof, use t = 4k + 1 and 4k + 3. (Why? Why not 4k + 2?). But help them to see the need for (ii) by using several values of t in (i).
6. Get the students to write up what they have discovered.

#### Extension to the problem

If the Post Office had 4c and tc stamps, and every amount of postage could be made from a given point on, what is the smallest value of that given point?

#### Solution to the problem

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

Let t be the value of the unknown denomination. The simplest way to do this problem is to guess and check. So you might guess t = 10. But you can’t get any odd numbers with 4 and 10. Guess t = 11. But 41 = 2 x 4 + 3 x 11. So that guess is incorrect. By gradually increasing the value of t you could discover that t is probably 15. Then you need to justify that 15 does not give 41 but does give everything from 42 onwards using the proof method of 4c and 7c Stamps,  Number, Level 5. (It’s necessary to do this otherwise we can’t be sure that t = 15 fits all the data of the original problem.)

A nicer method than the one above, is to know that in general the lowest ‘point’ is 3(t – 1). This can then be equated to 42 and the linear equation solved to give t = 15. (However, we don’t justify this lowest point until the Extension.)

#### Solution to the extension:

With 4 and t we can make 3(t – 1) and everything from then on but we can’t make 3(t – 1) – 1. The value of 3(t – 1) has to be guessed by experimenting with various values of t. (Note that this was encouraged in the Teaching Sequence above). But we have to be careful to remember that 4 and t have no factors in common (see 4c and 7c Stamps,Number, Level 5).

Proof: Using the proof method of 4c and 7c Stamps, Number, Level 5, we will need to show that we can get 3(t – 1), 3(t – 1) + 1, 3(t – 1) + 2 and 3(t – 1) + 3. Then we need to show that we can’t get 3(t – 1) – 1.

3(t – 1). Now we have to get this as a linear combination of 4 and t. But with 3(t – 1) = 3t – 2, it isn’t obvious how to do this. Check with t = 5, 7, and a few other values to see if this gives us any insight. It does because it suggests that we should write t as 4k + 1, 4k + 2 or 4k + 3. (It can’t be 4k as 4 and t have no factors in common.) But mature reflection tells us that we don’t need to worry about 4k + 2 either as 4k + 2 and 4 have a common factor of 2.

If t = 4k + 1, then 3s – 3 = 12k = 4(3k). We can do this with only 4c stamps.

If t = 4k + 3, then 3s – 3 = 12k + 9 – 3 = 12k + 6 = 4k + 2(4k + 3) = 4k + 2t. We can do this with 4c and tc stamps.

3(t – 1) + 1. To do this we need to follow in the footsteps of 3(t – 1).

If t = 4k + 1, then 3(t – 1) + 1 = 3t – 2 = 12k + 3 – 2 = 12k + 1 = 4(2k) + (4k + 1).

If t = 4k + 3, then 3(t – 1) + 1 = 3t – 2 = 12k + 9 – 2 = 12k + 7 = 4(2k + 1) + (4k + 3).

3(t – 1) + 2 = 3t – 3 + 2 = 3t – 1.

If t = 4k + 1, then 3t – 1 = 12k + 3 – 1 = 12k + 2 = 4(k) + 2(4k + 1).

If t = 4k + 3, then 3t – 1 = 12k + 9 – 1 = 12k + 8 = 4(3k + 2).

3(t – 1) + 3 = 3t. So this is easy, we just use three tc stamps here.

But then there is the problem of why we can’t do 3(t – 1) – 1. How do we prove that something can’t happen? (Look at What is s?, Algebra, Level 6 as a model to follow.) One way is to assume that it does and show that this leads to something that is clearly false. This implies that the original assumption had to be wrong. So let us assume that there do exist numbers of stamps a and b such that 3(t – 1) – 1 = 4a + bt.

Rearranging we get 3t – bt = 4a + 4. So t(3 – b) = 4(a + 1). Now we know that t has no factors in common with 4. But 4 divides the right hand side of the equation, so 4 divides the left hand side. So 4 divides 3 – b. Since 3 – b has to be positive (the right hand side is), 3 – b is either 1 or 2. But 4 doesn’t divide 1 or 2, so we have our contradiction.

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WhatisT.pdf35.6 KB
WhatisTMaori.pdf41.41 KB

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