Trial and Improvement

I'm starting to think this may be the way to get students to tackle those questions that they leave blank because they can't remember what to do.

Having studied at an OU summer school I was astonished how many students there used Trial and Improvement to answer the maths problems we were given. This was across the board including geometric, algebraic and trigonometric problems. I have also found that when asking adults to help me gain evidence for my OU maths education assignments they also use Trial and Improvement to tackle the questions I present them with.

The next question is which kind of questions can be answered using Trial and Improvement?
Can all questions be answered in this way?
Can students adapt the method to answer very different types of question?

The idea is that the student guesses the answer then computes if that answer leads to a correct conclusion. If it does they are finished but if not they have to use some sort of understanding to guess again and thus get closer to the correct solution. I'm not sure how many of the GCSE questions can be answered by this technique and I'm not sure how successful pupils will be using it but it is perhaps the one technique they can learn that will enable them to answer the questions they usually leave blank and thus gain no marks for.

I'll have to do more work on this to see how widely it can be applied.

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.

 

 











Simultaneous Equations 2

I'm not sure why I didn't see this connection before but now I've had a chance to think about this it is obvious how to generate equations, with integer co-efficient,  passing through the same point.

Take the point (- 2, 6). I just have to set up the co-efficient's for the x and y terms and then find out what they sum to.

For instance I want one equation to involve 2x and 3y and the other to involve 3x and 2y.
So I write:

2x + 3y = 2(-2) + 3(6) = 14.

Giving me the  equation:

2x + 3y = 14.

The second one is then:

3x + 2y = 3(-2) + 2(6) = 6

Giving me the second equation:

3x + 2y = 6.

I can vary the signs involved and can produce an infinite variety of simultaneous equations. These can vary in difficulty depending on the skills I wish to develop in a student.

This is connected to Diophantine equations and I can't remember most of the stuff I've learnt about these. Still this is a practical application I never thought of.

Simultaneous Equations

I was thinking about how to produce simultaneous equation questions with 'nice' numbers. Then I realised that I only had one sure way of producing a question. For example:

• I pick an equation x + y = 4
• I know that perpendicular lines to this are such that the product of their gradients is -1. So required line is y - x = c
• Choose a point on my chosen line, (- 2,6).
• Required line has c = 6 - (- 2) = 8
• Giving the other line as y - x = 8.

But there are lots, possibly infinite, lines that pass through this point, i.e. (- 2, 6). How are they defined?

This is something I've never considered before. (Is this 'across the grain' thinking?) I want these equations to have whole number co-efficients. Now I need to investigate and see if I can find a way of classifying them.