'Explain fully how you arrived at your answer'

Eric Brown, University of Edinburgh

It is easy to add these words to a mathematics problem; often not so easy to respond to them satisfactorily. Three of the likely difficulties are:

1. The solution appears to be so 'obvious' that further explanation may seem superfluous.

2. A correct explanation may involve some logical nuance which, even if grasped, may not be clearly expressed.

3. A complete explanation may be genuinely beyond the level of the solvers and it becomes a matter of how close they can be expected to get.

Each of these will be illustrated by one of the problems which have been recently set in the Primary Division of the 'Mathematical Challenge' competition. The discussion will indicate how childrens' responses may be used to separate those who simply 'solve the problem' from those who 'solve it like a mathematician'!

Problem 1

Yvonne has two brothers, Alan and Daryl. The product of the ages of all three children is 144. Alan is 11 years older than Daryl. How old is Yvonne? Explain fully how you arrived at your answer.

All readers over a certain age will instantly associate 144 with. Our current P7s may have to think a little longer, but once they notice this then surely the ages must be 12, 12 and 1.

What more can be said?

(One child made the interesting deduction that Yvonne and Alan must have been twins!)

Well, there could be other possibilities; the true mathematician is not content with one solution - he or she wants to know if there are any more and, if not, why not?

In this case, we can list the 15 factors of 144, or list the 18 different factorisations into 3 factors. (Incidentally, this is a skill which is well worth developing - it can lead to the investigation of some very important properties of whole numbers.)

We can now see that no two factors differ by 11 other than 12 and 1, so the solution is unique. Full marks!

Problem 2

Eggs are placed in a 5 by 5 egg carton in such a way that there are at most two eggs in each row or column and at most two eggs in each of the two main diagonals (from the top left hand corner to the bottom right, and from the top right hand corner to the bottom left). What is the greatest number of eggs that can be placed in the carton? You should state the number and provide a diagram which clearly indicates the positions of the eggs. Explain fully why there cannot be more eggs than the number shown in your diagram.

Most children will draw 10 eggs in a carton, fulfilling the conditions of the problem, and then say something like 'I cannot put any more eggs in this diagram, because . . . The subtlety here is that the problem has not been solved yet - we must show that 10 eggs is the maximum possible in ANY diagram, not just the one we have drawn. Short of mind-reading, how can we tell if the solver has realised this?

Two indications were looked for:

(a) an explicit statement that there are five rows (or columns), each of which can hold no more that 2 eggs, or

(b) a display of several different, but correct diagrams with a statement that no more eggs can be added to any of them.

The diagrams illustrate solutions that got 3 out of 4, or 4 out of 4 using these criteria.

Solution 1

For this problem, I just wrote out several grids (as shown on the question sheet) and filled in the positions of the eggs with crosses. I then checked for flaws in the layout.

I think the greatest number of eggs that it is possible to place in the carton if there are no more that two on each row is ten. For a diagram, see the five grids above that are labelled '10'. There cannot be more than ten eggs in the carton because if there were, there would be more than two on at least one row.

4 marks

Solution 2

There are 5 rows and only 2 eggs are allowed to be placed in each row. So therefore the maximum number of eggs you can have is ten. The diagram above shows that ten can be fitted in without breaking the rules.

4 marks

Solution 3

The greatest number of eggs which you can have in the container is 10.

I think this is correct because if you add even one more egg anywhere in the egg box then there would be 3 eggs in one of the rows, columns or major diagonals. I also have 2 eggs in each row, column and main diagonal.
There is no right or wrong way to do this problem because it is just a matter of trial and error.

3 marks

Problem 3

To cover the Christmas present (shown alongside) with sticky backed paper, you will require six rectangular pieces. What are the dimensions of the smallest single rectangular piece of sticky backed paper from which you could cut out the six pieces with the minimum of wastage?

Explain clearly how you arrived at your answer.

To cover the Christmas present (shown alongside) with sticky backed paper, you will require six rectangular pieces. What are the dimensions of the smallest single rectangular piece of sticky backed paper from which you could cut out the six pieces with the minimum of wastage?

Explain clearly how you arrived at your answer.

It is not difficult to be convinced that the arrangement 'has to be' essentially that shown here, which uses a rectangle with a 'wastage' of .

How could this possibly be 'proved'?

If we assume that the 'minimal' rectangle must have integer dimensions, then we could factorise each number between 202 (the surface area of the cuboid) and 207 (= 9 × 23) and observe that the resulting rectangles are clearly impossible, This is surely a lot to expect from a pupil in P7!

Here is a 'challenge' for the grown-ups: 'Given a cuboid with integer dimensions, is it true that the smallest rectangle from which the six faces can be cut must itself have integer dimensions?' Explain your answer fully!

Note that the result is not true for an arbitrary set of such rectangles, as a counter-example can be provided. The author does not have a rigorous proof for the stated assumption of integer dimensions - responses to this challenge would be welcomed.