Little Magic Squares

The Problem

Tui has just discovered magic squares. She decided to make all of the magic squares that she could just using the numbers 1, 2 and 3. How many could she make?

It took her quite a while because she didn’t know that the sum of a magic square was always three times the number in the centre.

Teaching sequence

1. Talk about square ‘arrays’ of numbers like the ones in A Square of Circles. Ask the class if you can put numbers into these arrays so that the rows have the same sum; the columns have the same sum; all of the rows, columns and diagonals have the same sum.
2. Show them a magic square such as the one below.

   6	1	5
   3	4	5
   3	7	2

3. Get them to check that the rows all have the same sum (of 12); that the columns all have the same sum; and that the diagonals have the same sum.
4. Tell them that these things are called magic squares and that the sum of a magic square is the common sum of the rows, columns and diagonals.
5. Tell them Tui’s problem.
6. Ask them to go away in pairs and see how many magic squares they can find.
7. Get some of the pairs to report back. Can they prove that the arrays they have produced are magic squares?
8. As the Extension problem is not so different from the original problem, most of the class might be asked to try it.

Extension

How many magic squares would Tui have made if she had only the numbers 7, 8 and 9 to use?

Note: This problem can be done with any three consecutive numbers. So you could assign them whatever numbers you would like the students to practice on. You will only get the four answers.

Actually you can take it even further and use three consecutive even numbers or three consecutive odd numbers. Again you only get four magic squares.

You might like the students to try other three numbers to see how many magic squares that they can find. You could find more or you may find less.

Solution

We should say right at the start that we don’t expect the students to solve the problem the way we do below. We would expect the students to use guess and check and to stumble across the final set of answers. We have, however, done this problem very systematically so that you can see and be absolutely sure that Tui should have found only 4 magic squares.

However, it might be worthwhile trying to lead the class into seeing that there is a systematic way of getting the four answers.

Being systematic in this problem could mean choosing different numbers for the centre square. So the centre square could be 1, 2 or 3.

centre square = 1: This means that the sum of the magic square has to be 3. This sum can only be made by using three 1s. So this magic square consists of all 1s. Call this square A.

centre square = 2: This means that the sum of the magic square has to be 6. Now 6 can only be made with 1 + 2 + 3 or 2 + 2 + 2. One way to get a magic square here is for all of the entries to be 2. This is not very interesting but it does give another magic square that we will call B.

Now suppose that the centre square (2) is used with 1 and 3 somewhere to get the sum of 6. Because of the symmetry of the square, we can assume without loss of generality that this is either done on the main diagonal or on the vertical column through the centre.

In the first case, the middle square in the top row is either a 2 or a 3. (It can’t be 1 because then the row sum would not be 6.) We follow through these two situations.

In the ‘2‘ case, we have to have a 3 in the top right-hand square. But then the last column can’t sum to 6.

In the ‘3’ case, the 2 in the top row and the 1 in the middle column are forced. This then means that there has to be a 1 in the middle square of the last column. This forces the 3 and 2 in the first column. A quick check shows that we have another magic square. Call this magic square C.

Now we have to worry about the 1, 2, 3 being in the centre row. Because of the symmetry of the square, we can assume that there is a 2 in the top left-hand square and a 3 in the top right-hand square. This forces the two 1s as shown and then the final 3 falls into place. A final check shows that this is a magic square.

The funny thing is that if we rotate this last magic square through 90° , then it looks exactly the same as C. So we don’t get a new magic square this way.

centre square = 3: This means that the sum of the magic square has to be 9. This can only be done if the three numbers that make up a row or a column are all 3s. So we get another uninteresting magic square that we will call D.

Tui should have found four magic squares. These are the ones we have called A, B, C and D.

Solution to the Extension

Tui would have found only four magic squares this time too. Using exactly the same method as before she would have come up with the following answers.