It is helpful if students have worked on the problem What is s?, Algebra, Level 6, before attempting this problem.
In this problem, students are given the amounts that can and can't be made with two denominations of postage, and are asked to find the value of those denominations. This is an example of an inverse problem. As these occur often, it is a useful skill to be able to solve problems from all angles.
Guided guessing and checking is one approach that students may use, however the Extension problem also involves the constructing of proofs. Students are likely to need support here as this construction may prove challenging. However, this is an important aspect of this problem, as it will develop mathematical skills of conjecturing (guessing), proving and using algebra. There is more than one solution to this problem.
Other problems in this series include $3 and $5 Stamps, Number, Level 3, $4 and $7 Stamps, Number, Level 5, $5 and $9 Stamps, Number, Level 5, What is s?, and What is t?, Algebra, Level 6. However proof and generalisations only become evident at this level (Level 6).
The Otehaihai Post shop has two denominations of stamps for big parcels. It can make up any amount of postage from $54 onwards. It can’t make up $53. What are the values of the denominations?
- 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.
- Have the students work on the problem with a partner.
- If the student struggle, 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).
- Share students’ solutions with the whole class. Note any different answers and any different methods of solution.
- Have as many students as possible attempt the Extension.
- Allow time for the students to write up what they have discovered.
Extension to the problem
Write 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?
The method of proof here is essentially that of What is s?, Algebra, Level 6.
This can be done by guessing and checking. It should produce four answers (see below).
A prefered method 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. 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.
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.