ALG 09

Week 9: As with the Week 7 tasks, we consider situations here involving several operations, where order matters. Pupils can approach the work on several levels:

• empirically, where they might simply notice that changing the order of the numbers being operated on changes the result;
• intuitively, where they develop some sense of why the result changes (eg, because we operate on some numbers more than on others);
• analytically, where they observe closely what happens to the numbers step by step.

Monday: Here we introduce pupils to the 'merge machine' that features in each of this week's tasks. What's going?

You might well be familiar with number pyramids, where numbers are placed in cells arranged in layers, and a where a given number is the sum of the two numbers immediately below it, as here:

Marge's merge machine is rather like an upside down pyramid, where a number is the mean (rather than merely the sum) of the two numbers immediately above it.
As with pyramids, changing the order of the numbers can change the final result.
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Tuesday: Having had the opportunity to explore the merge machine in Monday's task, we ask more focussed questions about particular results and a general rule.

The largest result is obtained when the largest number, in this case 20, is placed in the middle cell, ie for these two arrangements:
2, 20, 10 and 10, 20, 2.
Similarly, the smallest result occurs when the smallest number, in this case 2, is in the middle cell, ie for 10, 2, 20 and 20, 2, 10.
Pupils might well have thought initially that there would be only one arrangement for each outcome. However, from a brief consideration of the symmetry of the merge machine, it is clear that there must be two.
The fact that the largest result occurs when the largest number is in the middle cell (and vice versa) might help pupils gain a sense of what the machine is doing - somehow the middle number is 'reduced' less than the outer numbers. What precisely is going on? We consider a detailed method of analysis in Friday's task...
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Wednesday: Here we modify Marge's machine in a way that reduces the symmetry and that should help pupils get a better sense of what is happening to the numbers.

Pupils should discover that order still matters here. They might also realise that the crucial starting position is now the right-hand cell rather than the middle cell.
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Thursday: Here we ask more focussed questions again, concerning the size of the result for the given numbers and in general.

The largest result is obtained when the largest number, in this case 20, is placed in the right-hand cell, ie for these two arrangements: 2, 10, 20 and 10, 2, 20.
Similarly, the smallest result is obtained when the smallest number, in this case 2, is placed in the right-hand cell, ie for these two arrangements: 20, 10, 2 and 10, 20, 2.
The fact that there are two arangments in each case derives from the symmetry of the left side of the machine. The crucial role played by the right-hand cell, namely that the number it contains is not 'reduced' as much (or as often) as the other numbers, might again help pupils gain insight into what is going on.
Interestingly, the largest and smallest results here (13 and 8½) are the same as for Marge's original machine. Are the two machines equivalent??
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Friday: Here we look in detail at what happens to the given numbers at each step. How does Mark's way of writing the results express the fact that numbers are being 'merged'?

At each step, the machine takes the mean of a pair of numbers. We usually think of this process as one of adding the two numbers, then halving the result. However, for this task it is more illuminating to see the process as one of halving the numbers first, then finding the sum. It is even more illuminating if we don't actually perform the sum but leave it as an expression. Thus, for the given arrangement 2, 10, 20, pupils might be able to apprehend that 2 and 10 are in effect halved twice, while 20 is only halved once and so reduced by a smaller factor.
If we apply this form of analysis to Marge's original machine, for example to the arrangement 2, 10, 20 shown in Monday's task, we can see that the outer numbers 2 and 20 are halved twice, but that when the middle number 10 is first halved, this result is placed in two cells before being halved again. Thus we end up with 2/4 + 10/4 + 10/4 + 20/4, or 2/4 + 10/2 + 20/4. This is the same result as we would get on Mark's modified machine for the arrangment 2, 20, 10. So the two machines are equivalent, with the middle cell on Marge's version corresponding to the last cell on Mark's. [Note: we are not suggesting that you should take the analysis this far with any given class, but it might resonate with some pupils, as I hope it does with us!]