At it's simplest this problem involves the students adding single digit numbers. However as the students work towards finding all possible solutions they are involved in reasoning with these numbers. They may also use an understanding of symmetry (reflective and rotational) to show that some answers are in fact the same.
In the diagram there are 6 circles arranged in the shape of an equilateral triangle. You are given the numbers 1 to 6. Put a different number in each circle. How many ways are there of doing this so that the sums of the numbers on each side of the triangle are the same?
- Introduce the problem to the class. Pose the problem as a game or puzzle.
- Brainstorm ideas for approaching the problem and keeping track of answers. Have available the digit labels for those students who choose to move instead of write the digits.
- As the students work on the problem in pairs you may ask the following questions to extend their thinking:
What strategies helped you find the answer?
How can you use you knowledge about numbers here?
Are any of your answers similar – in what way? Are they different answers?
How did you know when you had the solution and could stop looking?
Can you think of a related problem to explore?
- Share answers – write them on the board as they are given. Discuss those answers that are reflections or rotations. Ask the students to explain their reasoning for why there are only 4 different answers.
Extension to the problem
What other six numbers can be used to make equal sums for each side?
The students will probably very quickly discover two things. One, that some answers can be obtained from others by reflecting or rotating the triangle. Two, that there are only four different ways to do this, subject to reflections and rotations.
The difficult part is to show that there are only four different answers. In order to do this you can first show that there are only four possible sums that lie between 9 and 12 inclusive. The reasons for this are (i) that 1 has to be somewhere; (ii) that the biggest sum that can be made using 1 is 1 + 5 + 6 = 12; and (iii) that 6 has to be somewhere; (iv) the smallest sum that can be made using 6 is 1 + 2 + 6 = 9. (This can be discovered in a number of other ways.)
So why is there only one arrangement with a sum of 9? How can you make up 9? There are only three ways: 1 + 2 + 6, 1 + 3 + 5, 2 + 3 + 4. These can then each be fitted into the three different sides. A similar argument can be used for sums of 10, 11, 12.
It’s important to try this last part so the students see how you can reason mathematically. How do you get to the bottom of something that appears quite complicated? You may want to do this as a whole class discussion.
The approach taken above is not the only way to do the problem. Explore the different methods that your students may suggest.