Fractional Magic Squares

Number and Algebra, Level 4

The Problem

Tui has really begun to get the idea of magic squares. She decided to make all of the magic squares that she could using the fractions 7/6, 4/3 3/2. How many can 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.

What is the problem about?

First of all, if the class hasn’t heard of magic squares, then you may need to tell them that a magic square is an arrangement like the one below where the vertical, horizontal and diagonal lines of numbers all add up to the same value. This ‘same value’ is called the sum of the magic square.

4 1 7
7 4 1
1 7 4

Magic squares are interesting objects in both mathematics proper and in recreational mathematics. So they are objects that students should have heard about and experienced. The problems in this sequence give students the opportunity to use the new numerical or algebraic concepts that they will have acquired at that Level, along with magic squares.

It’s a critical part of this and some earlier problems that three times the centre square is equal to the sum of the magic square. We prove this in the Extension to Negative Magic Squares, Level 4 and in the Level 5 lesson (The Magic Square).

This problem is in the sequence of problems on magic squares. The first of these is A Square of Circles (at Level 2), and no attempt is made to actually explore magic square properties there. The second lesson is Little Magic Squares (Level 2). There are essentially two magic square problems at Level 3 – Big Magic Squares and Decimal Magic Squares.

As well as this lesson, at Level 4, Negative Magic Squares uses negative numbers. This is followed by The Magic Square, Level 5. Finally, Difference Magic Squares at Level 6, looks at an interesting variation of the magic square concept.

Students at Level 5 or 6 might like to play with the consecutive numbers b – 1, b and b + 1 to show that there are only four magic squares that use only these three numbers. Does the same argument hold with 2b – 2, 2b and 2b + 2? What about b, c and 2c - b? Can you think of any other variations?

Relevant Achievement Objectives

Specific Learning Outcomes

The students will be able to:

Resources

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
  1. 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.
  2. 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.
  3. Tell them Tui’s problem.
  4. Ask them to go away in pairs and see how many magic squares they can find.
  5. Get some of the pairs to report back. Can they prove that the arrays they have produced are magic squares?
  6. How many different magic squares can they find? How many do they think there are?
  7. Ask the students to write up what they have discovered.
  8. As the first part of 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

Can you make up a magic square using any of the fractions 1/12, 1/6, 1/4, 1/3, 5/12, 1/2, and 7/12, with 1/6, 1/3 and 1/2, in that order, down the main diagonal?

Can you make up a magic square that has fractions in it?

Solution

There are actually four answers here if you allow the same fraction to be used in each entry of the squares. The neat way to see this is to notice that 7/6, 4/3, 3/2 is the same as 7/6, 8/6, 9/6. So now repeat the arguments of Little Magic Squares with 7, 8, 9 replaced by 7/6, 8/6 9/6.

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 = 7/6: This means that the sum of the magic square has to be 21/6. This sum can only be made by using three 7/6s. So this magic square consists of all 7/6s. Call this square A.

7/6 7/6 7/6
7/6 7/6 7/6
7/6 7/6 7/6

centre square = 4/3 (or 8/6): This means that the sum of the magic square has to be 24/6. Now 24/6 can only be made with 7/6 + 8/6 + 9/6 or 8/6 +8/6 + 8/6. One way to get a magic square here is for all of the entries to be 8/6. This is not very interesting but it does give another magic square that we will call B.

8/6 8/6 8/6
8/6 8/6 8/6
8/6 8/6 8/6

Now suppose that the centre square (8/6) is used with 7/6 and 9/6 somewhere to get the sum of 24/6. 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 8/6 or a 9/6. (It can’t be 7/6 because then the row sum would not be 24/6.) We follow through these two situations.

In the ‘8/6‘ case, we have to have a 9/6 in the top right-hand square. But then the last column can’t sum to 24/6.

In the ‘9/6’ case, the 8/6 in the top row and the 7/6 in the middle column are forced. This then means that there has to be a 7/6 in the middle square of the last column. This forces the 9/6 and 8/6 in the first column. A quick check shows that we have another magic square. Call this magic square C.

7/6 9/6 8/6
9/6 8/6 7/6
8/6 7/6 9/6

Now we have to worry about the 7/6, 8/6, 9/6 being in the centre row. Because of the symmetry of the square, we can assume that there is a 8/6 in the top left-hand square and a 9/6 in the top right-hand square. This forces the two 7/6s as shown and then the final 9/6 falls into place. A final check shows that this is a magic square.

8/6 7/6 9/6
9/6 8/6 7/6
7/6 9/6 8/6

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 = 9/6: This means that the sum of the magic square has to be 27/6. This can only be done if the three numbers that make up a row or a column are all 9/6s. So we get another uninteresting magic square that we will call D.

9/6 9/6 9/6
9/6 9/6 9/6
9/6 9/6 9/6

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

Solution to the Extension

Using the techniques that should have become standard if you have done the earlier magic square problems, you should find two magic squares. First note that the sum here is 1. The magic squares are:

1/6 1/2 1/3 1/6 5/12 5/12
1/2 1/3 1/6 7/12 1/3 1/12
1/3 1/6 1/2 1/4 1/4 1/2

You can get these by putting in the entries along the main diagonal and then trying each of the fractions in the centre entry of the top row. By finding the fraction that makes the row (column/diagonal) sum to 1 you will either complete the magic square or show that some other number is need to do so.

One way to cheat is to make up a magic square using whole numbers and then divide them all by the same number. The resulting fractions will make a magic square. You can see we did that with 7/6, 4/3 and 3/2.