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 one stamp denomination and the smallest postal value, and asked to find the other denomination. Guess and check 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.
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 s and t?, Algebra, Level 6. However proof and generalisations only become evident at this level (Level 6).
The Otehaihai Post Shop only sells two denominations of stamps for larger letters and for parcels. It has $4 stamps and stamps of some other value. It can make up any amount of postage from $42 onwards. It can’t make up $41. What is the value of the other denomination?
- Discuss mail. Ask:
How often do you or your family post letters and parcels?
Why have postage stamps become more expensive?
Who thinks there will still be letter mail in 5 years? Why do you think that?
- 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.
- Have the students work on the problem with a partner.
- If students struggle suggest that they experiment with various values of t (and keep a record of their work as it will be useful later).
- Share students’ solutions with the whole class. Note any different methods of solution.
- 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).
- Allow time for the students to write up what they have discovered.
Extension to the problem
If the Post shop has $4 and t$ stamps, and every amount of postage can 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 $4 and $7 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 preferred method 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). We have to be careful to remember that 4 and t have no factors in common (see $4and $7 Stamps,Number, Level 5).
Proof: Using the proof method of $4 and $7 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 this must become 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.) It is evident 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 $4 stamps.
If t = 4k + 3, then 3s – 3 = 12k + 9 – 3 = 12k + 6 = 4k + 2(4k + 3) = 4k + 2t. We can do this with $4 and t$ 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 t$ 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.