This problem involves the use of sums and multiples of 3 and 5. It also involves understanding that if you have 3 consecutive numbers then you can produce all subsequent numbers by adding multiples of 3 to those numbers. For example: starting with 13, 14, 15 and adding 3 to each gives 16, 17, 18 and then adding 3 to these gives 19, 20, 21 and so on.
This problem is the first of a series of six problems that appear at other levels. There is also a series of three Table problems students might explore before solving this problem. These are Jim’s Table, Algebra, Level 1, Jo’s Table, Algebra, Level 2, Sara’s Table, Algebra, Level 3.
The Otehaihai Post Shop only sells $3 and $5 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 (3, 5, 3 + 5 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 3 and 5 or not. You might suggest that they start at 1, then 2 and so on to see which values they can make.
- If they can see that it looks like everything from 8 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 3 to a single number or a set of numbers.
- Share solutions including 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 $3 and $7?
Can you investigate and suggest a general result with two denominations of stamps where one denomination is $3?
One approach is to make a table showing the numbers 1 to 20 and put a tick against those that can be made and a cross against those that can’t. From this it seems that 3, 5, 6, and everything from 8 onwards can be made. (See also Sara’s Table, Algebra, Level 3.)
This is a conjecture that should be justified: You can make 8, 9, and 10. Add 3 to each of these and the sums are 11, 12, and 13. Add 3 to each of these and the sums are 14, 15, and 16. In this way you will get any number you want that is bigger than 8, simply by adding enough threes.
Alternately, though a little longer, show that 8, 9, 10, 11, 12 can be made, and fives can be added to get to any number above.
Solution to the Extension:
The same approach will work with 3 and 7. Here you can get 3, 6, 7, 9, 10, and everything from 12 onwards.