The purpose of this activity is for students to apply additive and multiplicative thinking to find three addends with the same sum.
The background knowledge presumed for this task is outlined in the diagram below:
This activity should be used in a ‘free exploration’ way with an expectation that students will justify the solutions that they find.
Set of dominoes or printable dominoes
The procedural approach (show more)
- The student calculates effectively to find if each side has the same sum. .
A procedural approach involves trying various combinations of dominoes to obtain a combination that gives the same side total. Usually students use the following procedure. Choose a starting domino and build onto that domino so the side totals are equal.
Choose a third domino to make the same side total than check if the fourth domino can be placed.
Students might also decide on a side total and attempt to find dominoes that will work. They will need to adopt a conceptual approach to finding the possible donuts.
The conceptual approach (show more)
- The student uses knowledge of all the dominoes in a six-dot set to find dominoes to match a given side total.
The set of possible dots is all the paired combinations of the numbers 0, 1, 2, 3, 4, 5, 6.
For a given side total a conceptual approach might go like this:
Take a side total of 12. That means that all four sides total 48 but the corners are counted twice.
Suppose the numbers 2, 3, 4, and 5 are selected to be the corner sets of dots. Those numbers total 14. Will that work? Subtracting 28 from 48 allows for the double counting of corners and gives 20. That means the numbers that go into the central position on each side total 20. Not all arrangements of 2, 3, 4, and 5 in the corners will work.
This approach can be applied to find the donut with the highest side total. The dominoes 6-6, 6-5, 5-5 and 6-4 have the greatest number of dots, 43 in total. To allow for double counting at the corners a number must be added to 43 to make a multiple of four. The highest possible corner numbers are 6, 6, 5, and 6 which equals 23. 43 + 23 = 66 which is not a multiple of four. However, 64 is a multiple of four so 6, 6, 4 and 5 may work with a side total of 64. Each side has 64 ÷ 4 = 16 dots. Playing around with the arrangement of 6, 6, 4 and 5 in the corners produces this donut that has the highest possible side total.
Similar thinking can be used to find if a side total of two is possible. That would give a total of eight with the corner numbers counted twice. The dominoes with the smallest number of dots are 0-0, 0-1, 1-1, 0-2. That is a total of five. Since the next multiple of four is eight, having three dots on the corners might work. The only way to do that is to have 0, 0, 1, and 2 at the corners. Playing around with arranging those numbers in the corners gives this donut.