Abigail who goes to Histon and Impington Infants School sent us
in a very detailed solution to this problem. She wrote:
We named the squares with letters and then we wrote down the three
rules in letters:
A B
C
D E
1. AxB=C
2. D+E=C
3. B+C+D=A+C+E
I then made a guess, that if A=6 and if B=4, C must be 24, but this is too big, so B must be smaller. I tried B=2 which is still too big, then B=1, but then C=6, but this is not different to A, so A cannot be 6.
I guessed that A might be 1 and B=2, but then C = 2 too. I saw that A can't be 1 because then B and C would be the same.
I guessed that A might be 5, and tried B=2, but then C is too big, but B can't be 1 because A would be the same as C so I tried another value for A.
I guessed that A might be 3, and B =2 then C=6 which is possible. I tried other B values.. Not 1, not 3, maybe 4, but that is too big.
Then I guessed A might be 2. If A = 2 and B=3, C=6 is
possible.
If A= 2 and B=4, C=8 is possible. B can't be 5 because C would have
to be 10. B can't be bigger than 5 when A is 2.
I then wrote down what I had found so far.
A = 6 is not possible
A = 1 is not possible
A = 5 is not possible
A = 3 is possible
A = 2 is possible
We could try A = 4, but we don't need to try bigger than 6, because then C would be too big.
So if A = 4. B could be 2 and C could be 8. But if B is 3, C is too big.
With all of the possible answers, C is either 6 or 8
Then I looked at D + E = C.
If C = 6 I wrote a table:
D E
1 5
2 4
3 3
4 2
5 1
0 6
6 0
D or E can't be 2 because 2 is already used in A or B. They can't be 3 because they need to be different. They can't be 6 because that is C.
If C = 8, D and E could be...
D E
0 8 but C=8
1 7
2 6 but A or B = 2
5 3
4 4 but they can't be the same
3 5
6 2 but A or B = 2
7 1
8 0 but C =8
Then Mummy helped me write a big table with all the possible answers, and I worked out the sums for rule 3:
A B C D E B+C+D A+C+E
4 2 8 1 7 11 19
4 2 8 5 3 15 15 OK
4 2 8 3 5 13 17
4 2 8 7 1 17 13
2 4 8 1 7 13 17
2 4 8 5 3 17 13
2 4 8 3 5 15 15 OK
2 4 8 7 1 19 11
2 3 6 1 5 10 13
2 3 6 5 1 14 9
3 2 6 1 5 9 14
3 2 6 5 1 13 10
So the answers are:
4 2
 8
5 3
and
2 4
 8
3 5
When I looked at these answers, I noticed that they are mirror images.
This is fantastic reasoning, Abigail. You have really gone
through the problem in a systematic way. Well done!
Thank you also to Katie, Kirsty, Sam, Lucija, Marek, Rosie, Sarah,
Natasha, Nathaniel and William who told us they had each found one
of the solutions.