Name: Helen

Who is asking: Teacher
Level: Secondary

Question:
I am currently a student teacher in the UK and I have to write a 1000 word report on the following maths problem which I am completely stuck on! PLease HELP!! Choose and 3 by 3 section of the hundred square. Add the total of the four corners. How many different groups of four numbers can you find that add up to that number? eg,

1	2	3
11	12	13
21	22	23

Total of 4 corners add up to 48.

Adding 2, 13, 22, 11 also make 48 etc..

How many different groups of 4 numbers would add up to 48?

How would these results compare with thoses obtained from a 3 by 3 square in which the numbers are consective? eg,

1	2	3
4	5	6
7	8	9

PLEASE HELP AS I AM COMPLETELY STUCK? WHY DO ALL THESE DIFFERENT WAYS ADD UP TO THE SAME NUMBER??

Hi Helen

I think the way to simplify this problem is to do it algebraically. Suppose that the number in the upper left corner of your 3 by 3 section of the hundred square is a, then this square can be written

a	a+1	a+2
a+k	a+k+1	a+k+2
a+2k	a+2k+1	a+2k+2

In the example that you gave a=1 and k=10. All that you know about k in general is that it is at least 3, and in fact if k=3 then you are in a situation where the numbers are consecutive.
   If you add the four corners you get 4a+4k+4. Notice also that if you add any four numbers in the array you get 4a+something, and hence to see if you get 4a+4k+4 you can ignore the a's and see if what remains gives a sum of 4k+4. Thus you only need consider

0	1	2
k	k+1	k+2
2k	2k+1	2k+2

where the sum of the four corners is 4k+4.
   You now need to go through this array and see how to choose 4 entries and get a sum of 4k+4. Notice that you must have at least one of the entries in the last row or you can't get 4 k's, and you can't have all three of the entries on the last row or the sum has 6 k's. I think that you now need to go through the possibilities in an organized way. For example if you choose only 2k from the last row what entries from the first two rows will give a sum of 2k+4? Now do the same with 2k+1 and 2k+2. After you have done this see what hapens when you choose two entries from the last row. Notice that in this case the remaining two entries must come from the first row.

Good luck,
Harley