Difference Magic Squares
This is a problem from the number and algebra strand.
Use algebra when necessary
Construct, 3 by 3 arrays of numbers according to a given rule
Devise and use problem solving strategies to explore situations mathematically (be systematic).
This problem is the eighth and final lesson in a sequence of problems on magic squares. The first of these is A Square of Circles (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.
At Level 4, we have Negative Magic Squares that uses negative numbers and Fractional Magic Squares that uses fractions. Then The Magic Squares at Level 5, looks at a commonly seen magic square that uses just the numbers 1 through 9.
This Difference Magic Square can be solved by guess and check, where the students guess where the numbers go into the 3 by 3 grid and then check whether this gives the correct solution. The most efficient way to solve this problem, however, is to use algebra. It would be a good idea to have your students first try non-algebraic techniques, such as guess and check, to emphasise the efficiency of algebra.
Tui has become pretty well exhausted by magic squares. But she decides to have one more try but with a twist.
She wonders if it is possible to place the numbers 1 through to 9 in a 3 by 3 grid, so that in any direction of three squares (across, down or diagonally) the sum of the first and last numbers minus the centre number gives the same answer.
If it is possible, how many different answers exist?
Write the problem on the board, and read it to the class.
Check that the class understands the problem.
Ask them what the problem’s important information is, and write this on the board.
Break the class into groups and let them play with the problem.
Move from group to group and establish what progress they are making.
Have you seen a similar problem before? How did you approach that? Could you apply equations to the problem?
Reconvene the class and have them discuss their progress so far. Write on the board the steps made to that point.
Break into groups again and have the students explore the problem further.
Continue repeating sequences 4-7 until the problem is solved
Give the Extension problem for homework (see below). Have students show their working.
Let the students write up what they have discovered.
Drawing up a 3 by 3 grid, label the grid cells as following:
Let k = the sum of the first and last numbers minus the centre number. Therefore:
k = A + C – B = G + I – H = B + H – E = A + I – E
= D + F – E = A + G – D = C + I – F = C + G – E
We can see that:
k – E = D + F = B + H = A + I = C + G.
We know that:
A + B + C + D + E + F + G + H + I = 1 + 2 +…+ 9 = 45.
Rearranging we get:
(A + I) + (B + H) + (C + G) + (D + F) + E = 45.
4(A + I) = 45 – E. Thus: A + I = 45 - E .
Trying 1 to 9 as values for E, only E = 1, 5, and 9 give whole number answers for A + I:
This gives the following possible combinations of A and I.
Many 3 by 3 designs could be constructed from these combinations. Perhaps we could limit the number of designs further. Looking back at the original equations, is there another equality we can find involving E, A and I from the four equations we have not yet utilised?
k = A + C – B = G + I – H
= A + G – D = C + I – F
We can see that:
(A + C – B) + (G + I – H) + (A + G – D) + (C + I – F) = 4k.
Rearranging we get:
2A + 2C + 2G + 2I – (B + D + F + H) = 4k.
We previously established that k – E = D + F = B + H = A + I = C + G.
4(A + I) – 2(A + I) = 4(A + I – E).
This reduces to:
2E = A + I.
The only E, A, I combinations that fit both A + I = (45 – E)/4 and 2E = A + I are those that have E = 5. When E = 5, A + I = 2E = 10, thus k = A + I – E = 10 – 5 = 5 = E.
Using this knowledge, and the equation A + C – B = 5, when E = 5,
if A = 1 and I = 9, then:
C – B = 4.
Of the remaining numbers, B and C could be: 2 and 6, 3 and 7, 4 and 8.
If C = 6 then:
C + I - F = 5, making F = 10, - not possible. So C cannot be 6.
If C = 7 then F = 11, again not possible. C has to be 8. But this gives F = 12. A contradiction, therefore A and I cannot be 1 and 9.
If A = 2 and I = 8, then C – B = 3. B and C could be: 1 and 4, 3 and 6, 4 and 7, 6 and 9. If C = 4, then F = 7.
As D + F = B + H = A + I = C + G = 2E = 10, then knowing F = 7, D = 3, G = 6 and H = 9.
We have found a solution!
(Due to the symmetry of a 3 by 3 grid, reflecting or rotating it can form 7 different arrangements of this solution.)
Now we need to establish if there are any more possible solutions:
If A = 3 and I = 7, then C – B = 2. B and C could be: 2 and 4, 4 and 6, 6 and 8.
If C = 4, then F = 7 – not possible, as I = 7. So C cannot be 4.
If C = 6, then F = 8.
As D + F = B + H = A + I = C + G = 2E = 10, then knowing F = 8, D = 2, G = 4 – not possible, as B = 4.
So C cannot be 6. If C = 8, then F = 10 – not possible.
A contradiction, therefore A and I cannot be 3 and 7.
If A = 4 and I = 6, then C – B = 1. B and C could be: 1 and 2, 2 and 3, 7 and 8, 8 and 9.
If C = 2, then F = 3. As D + F = B + H = A + I = C + G = 2E = 10,
then knowing F = 3, D = 7, G = 8, and H = 9.
However, this is simply a reflection of the previous solution. Therefore, there is only the one solution to this problem, which can be arranged in any of the following eight ways: