magic squares

these task sheets (and solutions) can be clicked to produce and save larger images
easier tasks are in the older posts and become more demanding towards more recent posts
hopefully the resources illustrate that 'magic' squares provide a context for a variety of skill practice - with:
- some form of problem solving requested;
- considerations about relationships, justification and proof;
- extending work to an involvement of symbols;
- developing to quite complex uses of algebra.

I am indebted to Martin Hansen, whose articles in Maths in School (march 2010, sept 2010 and and nov 2010) provided much clarity on a possible teaching sequence and an understanding of relationships and solution techniques

Monday, 22 August 2011

proofs 3

Prove that any three sets of three numbers will form a 3 by 3 magic square providing:
  • each set of three must be in a linear sequence with the same difference 'a' (e.g. 1, 7, 13 & 10, 16, 22 & 19, 25, 31 all with a common difference of 6)
  • the first terms in the three sequences must also be in a linear sequence, difference 'b' (1 , 10 and 19 in the above example)
this is why any 3 by 3 block from a calendar (with a = 1 and b = 7) will form a magic square













for any magic square
  • use the fact that the corners are the average of the two outer middles not in line with the corner (proved elsewhere) to establish that the lowest and highest numbers must be in a middle edge cell position
  • use the line total property to establish that the four corner numbers sum to a multiple of 4
  • what happens when (using the terminology above) b = 2a? try to explain why this happens

proofs 2

the 'Lo Shu'

this magic square consists of all the digits, from 1 to 9 inclusive

although there are a total of 8 rotations and reflections of it there is essentially one such magic square

what do 1 to 9 add up to?

what must each line total be?


think of some (good) reasons why the central number must be 5

by considering what numbers can be in the same line as 9, prove that 9 cannot occupy a corner position (and neither can 1)

by considering sums of odd and even numbers, prove that the corner numbers must all be even

given that 9 must be in a middle edge position, where must 8 go?