Big Magic Squares

The Problem

Tui has begun to like magic squares. She decided to make all of the magic squares that she could using the numbers 222, 555 and 888. How many could she make if she used each number at least once in the square?

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 to the problem

How many magic squares would Tui have made if she had only the numbers 777, 888 and 999 to use? (A given number can be used more than once or not at all.)

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. We would expect the students to use guess and check and to stumble across the final answer. We have, however, done this problem very systematically there so that you can see and be absolutely sure that Tui should have found only one magic square. (The other three that they might find use only one number and here we said that Tui had to use each number at least once.) It might be worthwhile trying to lead the class into seeing that there is a systematic way of getting the answer.

Being systematic in this problem could mean choosing different numbers for the centre square. So the centre square could be 222, 555 or 888.

centre square = 222: This means that the sum of the magic square has to be 666. This sum can only be made by using three 222s. So this magic square consists of all 222. However since we have to each number at least once this one isn’t a solution.

centre square = 555: This means that the sum of the magic square has to be 1665. Now 1665 can only be made with 222+555+888 or 555+555+555. One way to get a magic square here is for all of the entries to be 555. However since we have to each number at least once this one isn’t a solution.

Now suppose that the centre square (555) is used with 222 and 888 somewhere to get the sum of 1665. 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 555 or a 888. (It can’t be 222 because then the row sum would not be 1665.) We follow through these two situations.

In the ‘555‘ case, we have to have a 888 in the top right-hand square. But then the last column can’t sum to 1665.

In the ‘888’ case, the 555 in the top row and the 222 in the middle column are forced. This then means that there has to be a 222 in the middle square of the last column. This forces the 888 and 555 in the first column. A quick check shows that we have a magic square.

Now we have to worry about the 222, 555, 888 being in the centre row. Because of the symmetry of the square, we can assume that there is a 555 in the top left-hand square and a 888 in the top right-hand square. This forces the two 222s as shown and then the final 888 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 the last magic square. So we don’t get a new magic square this way.

centre square = 888: This means that the sum of the magic square has to be 2664. This can only be done if the three numbers that make up a row or a column are all 888s. However since we have to each number at least once this one isn’t a solution.

Therefore there is only one magic square solution to this problem.

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.