This problem involves the use of sums and multiples of 4 and 7 (linear combinations of 4 and 7). It also involves understanding that if you have 4 consecutive numbers then you can produce all subsequent numbers by adding multiples of 4 to those numbers. For example: starting with 23, 24, 25, 26 and adding 4 to each gives 27, 28, 29, 30, and then adding 4 to these gives 31, 32, 33, 34, and so on.
This problem is the second 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 4, 5c and 9c Stamps, Number, Level 4, What is s?, Algebra, Level 6, What is t?, Algebra, Level 6, and What is s and t?, Algebra, Level 6.
The Otehaihai Post Shop only sells $4 and $7 stamps for larger letters and for parcels. What amounts of postage can be made up from these denominations?
- Discuss mail. Ask:
How often do you or your family post letters?
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.
- As a class list some of the values (4, 7, 4 + 7 etc).
- Have students work on the problem making it clear that they are looking for some simple way of telling if a number can be made from a combination of 4 and 7 or not. You might suggest that they start at 1, then 2 and so on to see which values they can make.
- If they conjecture that it looks like everything from 18 onwards can be made, ask them if they can justify this. You might need to encourage them to think about what happens when you keep adding 4 to a single number or a set of numbers.
- Share solutions giving a justification for what they have found.
- Make the Extension problem available.
Extension to the problem
How would things change if the stamps were $4c and $9?
Can you guess a general result with two denominations of stamps where one denomination is $4?
The solution method follows the same pattern as that of 3c and 5c Stamps, Level 3.
A good way to start here is to experiment. For instance, make a table showing the numbers 1 to 30 and put a tick against those that can be made and a cross against those that can’t. What amounts seem to be working are 4, 7, 8, 11, 12, 14, 15, 16, and everything from 18 onwards.
How can this conjecture be justified? Can you make 18, 19, 20, and 21? Yes. Then add 4 to each of these and you’ll get 22, 23, 24, and 25. But then add 4 to all of these and you’ll get 26, 27, 28, and 29. Can you see now that eventually you will get any number you want that is bigger than 18, simply by adding enough fours?
(Alternately, though a little longer, show that 18, 19, 20, 21, 22, 23, 24 can be done and add sevens to get to any number above 18 that you want.)
Solution to the extension
The same approach will work with 4 and 9. Here you can get 4, 8, 9, 12, 13, 16, 17, 18, 20, 21, 22, and everything from 24 onwards.
Now 4 and t is different. We suggest you get the students to try various values of t so that they can look for patterns and produce a conjecture. Forget about the small values you can get. They might come up with the conjecture that, from 3(t – 1) onwards, you can get all numbers. That is on the track but doesn’t work if t = 8. In fact you need 4 and t to have no factors in common in order to get 3(t – 1).
Full proof of this last conjecture is given in What is t?, Level 6.