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.
Magic squares are interesting objects in both mathematics proper and in recreational mathematics. It is likely that students have already encountered magic squares. The problems in this sequence give students the opportunity to use known numerical or algebraic concepts.
It’s a critical part of this and some later problems that three times the centre square is equal to the sum of the magic square.
This problem is part of a series exploring magic squares. The first of these is Little Magic Squares and A Square of Circles at Level 2, Decimal Magic Squares also at Level 3. At Level 4, Negative Magic Squares, uses negative numbers and Fractional Magic Squares uses fractions. The Magic Square, Level 5 shows why three times the centre number is equal to the sum of the magic square. Finally, Difference Magic Squares at Level 6, looks at an interesting variation of the magic square concept.
Tui likes magic squares. She decides to make all of the magic squares that she can using the numbers 222, 555 and 888. How many can she make if she uses each number at least once in the square?
It takes her quite a while because she doesn’t know that the sum of a magic square is always three times the number in the centre.
- Talk about square ‘arrays’ of numbers like the ones in A Square of Circles. Ask the class if they 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.
- Show them a magic square such as the one below.
6 1 5 3 4 5 3 7 2
- Have them 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. This is why that are called magic squares.
- Pose Tui’s problem.
- Have them work individually or in pairs to see how many magic squares they can find.
- As solutions emerge, have some students share and prove that the arrays they have produced are magic squares.
- Pose the Extension problem as appropriate.
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.
You can take it 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.
Students are unlikely to solve the problem the way shown below. Students are more likely to use guess and check and to stumble across the final set of solutions. However, a systematic approach is shown here to show that Tui should have found only one magic square.
You may however want to have your students see that there is a systematic way of finding the solution.
Being systematic in this problem can 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.