"Tell me the next two numbers in each of these seven minor spells",
chanted the Mathemagician, "And the great spell will crumble away!"
Can you help Anna and David break the spell?
Exploring and predicting folding, cutting and punching holes and
There are seven pots of plants in a greenhouse. They have lost their labels. Perhaps you can help re-label them.
proved quite hard to explain. Chris G and Chris B also from
Moorfield Juniors had a good go:
In the end we worked out
the diagonals=the target number (100) because each across does:
and each down is:
I noticed that selecting
circled numbers in the way described gives one number in each row
and one number in each column. Combining these pieces of
information tells us that the total will always be the same.
It doesn't matter what
order we do the addition in, we'll end up with 21 + 23 + 15 + 30 +
0 + -4 + 5 + 10 = 100. It's very easy to construct other matrices
with this property now: just pick numbers for the row and column
headings that add up to the required total.
Bristol Grammar School also thought about this problem, and made
some interesting observations:
Let us label our matrix
rows A,B,C,D and columns 1,2,3,4, and let (A1) denote the value of
the top-left square.
In circling four numbers in
different rows and different columns, we select four squares Aa,
Bb, Cc, Dd where (a,b,c,d) is a permutation of (1,2,3,4).
In this solution I shall
prove the necessary conditions for the sum (Aa)+(Bb)+(Cc)+(Dd) to
be the same for all permutations (a,b,c,d). Let us say
(A1)+(B2)+(C3)+(D4) = n (so the sum on a diagonal is n) Now,
suppose we have a permutation where (a,b,c,d)=(2,1,3,4).
We can reduce this to the
above case by swapping the values of A1 and A2 and of B1 and B2 in
the matrix, so now all our values lie on the diagonal. Our
condition states that the sum must be the same, so if we consider
the 2x2 matrix (A1, A2)(B1, B2), we can see that our condition
holds only if the sums of the diagonals of this matrix are the
By repeated use of this
swapping system we can transform our matrix into one where the
circled values lie on our diagonal. Suppose a is not 1, then we can
swap Aa with A1 and X1 with Xa (where X is the row with the circle
in column 1). If b is not 2 we can swap Bb with B2 and Y2 with Yb
(where Y is the row with the circle in column 2). Note that Y
cannot be row A because we know its circle is now in column 1. We
apply this once more to row C (if c is not 3) to complete the
rearrangement. Note that, again, for this to actually work, it is
necessary that (A1)+(Xa)=(Ax)+(X1) (and similarly for rows B and
C), because we know that ultimately after the swapping is complete
X1-> Xa-> Xx is going to end up on the diagonal, since it is
Analysing this method of
transformation onto the diagonal, it is clear that our condition
will hold for all permutations if and only if for every rectangle
in our matrix (of size greater than or equal to 2 x 2), the sum of
the values in one pair of opposite corner squares is equal to the
sum of the values in the other pair of opposite corner squares.
This is a property
possessed by the two matrices given as examples, and any other
matrix with this property will also obey the condition. It is also
not necessary to restrict oneself to 4 x 4 matrices. Any square
matrix can exhibit the same behaviour (the proof is merely
induction on the aforementioned method).
explain it more clearly or add anything to this? If you think you
can then email us.