Simple Addition Problem

This piece of work looks at Question 18 of the UK Junior Mathematical Challenge April 2012.

 

 

 

The solution is easy to obtain:

 

 

 

 

 

How do we do this. Add up the numbers given , i.e. 2, 3, ..., 8. They total 35. The column and row have to add to 21, giving a total of 2 x 21 = 42. The difference between these two totals is

 

42 - 35 = 7

 

 

So 7 is added twice, hence it goes into the box marked X. We know each column/row adds to 21 and we already have 7 in place; leaving 21 - 7 = 14. By inspection we can find the two sets of available numbers which will add to 14, as seen in (1) above.

 

 

I wanted to look at this problem to see if there was an easy way to make my own questions based upon it. So I firstly simplified the diagram slightly so I could then see a way of producing similar problems.

 

 

 

 

There is also no need to use the numbers 2, 3, ..., 8. These can be altered to 1, 2, ..., 7 without loss of generality.

 

 

 

As can be seen this shows the same solution as the original problem except everything is reduced by 1.

 

 

Firstly I looked to see how many different solutions were possible. I used the diagram in (3) above and the numbers 1, 2,..., 7. These numbers total to 28. The total of the column and row must be an even number so I consider the totals 30, 32 and 34. With each total I found the difference between the two totals, i.e. 2, 4, 6. This gave me the numbers that go into the common box between the column and row. I then halved the totals, 30, 32 and 34 and subtracted 2, 4, and 6 respectively to get the remaining totals required to fill in the diagram. Inspection of the numbers then provided these solutions.

 

 

 

 

I then considered how to generalise this problem. I need the diagram to be a generalisation and the numbers used needed to be consecutive numbers. The first in the sequence of these diagrams would be a single box but that was too simple so I considered a diagram with 2 boxes in the column/row.

 

 

 

There is no way to fill this diagram. I have no idea how to prove this yet.

 

 

I then considered a diagram with 3 boxes in the column/row.

 

 

 

 

I also considered a diagram with 5 boxes in the column/row.

 

 

 

 

 

These all worked and I suspect larger diagrams will also work. Since the total of the column and row together is always even the shared box will be contain either an even or an odd number depending on whether the total obtained is even or odd respectively.

 

 

There are a number of questions I want to consider:

 

 

• Can I show that these are the only solutions?
• Why doesn't the diagram for 2 boxes in the column/row work?
• Can I prove the solution to one of these diagrams?
• Can I extend that proof to all diagrams?





 

I need to look at the problem a bit more to see how to develop a proof and answer the questions above.