This week we look at two tasks of classic form, both of which are rather contrived but which are also quite engaging if simply treated as a puzzles. The aim is to show the power of using a symbolic algebra approach, although in the case of the first task, below, we have deliberately devised a task which is also amenable to informal methods.

MONDAY: This first task is of the classic 'How many cows and chickens' form, although we have tried to make it a bit more challenging by involving three rather than just two sets of 'elements', in this case, bikes, trikes and quadracycles.
As we spellout in Tuesday's variant, this
Cycles task can be solved algebraically by creating equations such as
T = Q + 5
31 = B + T + Q
76 = 2B + 3T + 4Q.
We can then substitute Q for T. Also, if we double the second equation, giving 64 = 2B + 2T + 2Q, we can eliminate B by subtracting this equation from the third equation, leaving us with 14 = T+ 2Q.
In Tuesday's variant, we also indicate that the task can be solved informally with this sort of argument:
Imagine that the cycles are all bikes;
we would then have 64 tyres (or wheels) which is 14 fewer than the actual number;
so we would need to replace some of the bikes with trikes and/or quads;
each trike would increase the number of tyres by 1, each quad would increase the number of tyres by 2;
so ....
There is a clear link between this narrative approach and the process that gave us the equation 14 = T + 2Q. Of course, students might not see this link immediately, though Freudenthal (in his
China Lectures published in 1991) suggests that students can achieve this by solving a range of similar tasks in this informal way and generalising.
There is an interesting tension here. One of the strengths of a symbolic approach is that it 'frees' us from thinking about the context, once we have expressed the given information symbolically, and assuming we are fluent in manipulating the symbols.
On the other hand, the act of keeping hold of the context, or of referring back to it periodically, can help us (or more to the point, our students) to form the symbolisations and to check that these, and the subsequent manipulations, are valid.

TUESDAY: Here we present two methods for solving yesterday's
Cylces task, one informal the other involving symbolic algebra, which we ask students to complete.
How students respond to these methods will depend on their knowledge and experience. Some students might gain a better appreciation of the power and efficiency of symbolic algebra, while others might gain some understanding of the algebraic approach by linking it to the more concrete, informal approach.

WEDNESDAY: Here we present two new tasks, both set in the context of 'sharing acorns' but where the given information is structured differently. We consider how amenable the tasks are to informal and formal approaches in the light of this difference in structure.
The first task can be solved in informal and formal ways that are very similar: the collection of 60 acorns can be partitioned into 1 plus 4 plus 5 equal shares, ie into 10 equal shares altogether; algebraically, we can construct something like this:
g + 4
g + 5
g = 60, so 10
g = 60.
The second task
can be solved informally but it probably requires quite a lot of insight to do this successfully. On the other hand, it can be solved in a fairly routine way using symbolic algebra, assuming the solver has adequate technical skills.
Thus, we would argue that the second task is that elusive entity: a task (albeit a mere pointless puzzle!) where
the use of symbolic algebra comes into its own. Of course, in the first task, we could also strengthen the need to record and symbolise, by increasing the number of recipients and by relating their shares to George's share in more varied ways.
Note 1: You may have noticed that the methods of sharing in the two tasks turn out to be equivalent. The second task has been made more complex by using what Dettori et al (2006) call 'circular references'. I came across this
work in the excellent revue by Mason & Sutherland (2002),
Key Aspects of Teaching Algebra in Schools. You might like to devise similar tasks to challenge your students, by taking a straightforward sharing task and then using circular references to recast the description of some of the shares.
Note 2: The 'circular references' task used by Dettori (and reproduced in Mason & Sutherland) is this:
Interestingly, an almost identical task occurs in an English translation (1822, p204) of a French translation of Leonhard Euler's
Vollständige Anleitung Zur Algebra (1771).
The English book (below) contains 'Additions of M. De La Grange' and it
is possible that the problem stems from him rather than Euler, as I can't find it in the
original German edition.
THURSDAY: Here we introduce Auntie's way of sharing acorns. It is not equivalent to Grandpa's method, but it turns out to produce the same result for a particular number of acorns.
We can find the particular number of acorns using symbolic algebra, by solving the equation resulting from equating Peppa's shares (ie 4
g and 2
g + 20) or Chloe's shares (5
g and 2
g + 30) under the two ways of sharing. Both give the value
g = 10, so that the total number of acorns is 100.
We have obviously carefully contrived the two sharing methods for them to produce the same outcome for all three recipients for a particular number (100) of acorns. This wouldn't happen by chance. However it raises an interesting (although rather subtle) question: given that the choice of Grandpa's or Auntie's sharing method makes no difference for 100 acorns, does it make a difference, for any individual recipient, for other numbers of acorns?
FRIDAY: Having established that Grandpa's and Auntie's method happen to
produce identical shares when there are 100 acorns, today we consider
what happens when there are more than 100 acorns: for each individual
recipient, is one sharing method more favourable than the other?
Note: This task is quite complex and is probably more suitable as a challenge for teachers than as an activity in class.
In the case of George, he always gets one tenth of the acorns when Grandpa shares them out. This is not a large proportion, but still larger than what he gets from Auntie when there are less than 100 acorns. In the extreme case, when there are only 50 acorns, George gets none! George is better off with Auntie when there are more than 100 acorns.
The situation with Peppa turns out to be rather surprising. When there are 100 acorns, George's share is 10 acorns by either sharing method and she gets 40 acorns by either sharing method (ie 4×10 or 2×10 + 20). Now, when Georges share (
g) is greater than 10, 4
g will be greater than 2
g + 20, which suggests that Grandpa's method will be more favourable than Auntie's when there are more than 100 acorns. However, it turns out that this is not the case! Why? Because Peppa's share depends on George's (as it's a function of
g), but George's share differs depending on whether Grandpa or Auntie are sharing out the acorns. It turns out that Peppa's share comes to the same amount, whether it's Grandpa or Auntie doing the sharing, regardless of the number of acorns being shared out! For exmple, for 50 acorns she gets 20 under both methods, for 110 acorns she gets 44 under both methods. We leave it to the reader to verify/prove this.
For Chloe, it looks like Grandpa's method is more favourable than Auntie's when there are more than 100 acorns, as 5
g > 2
g + 30 when
g > 10. But can we be sure? In the case of Grandpa's method, Chloe's share is 5/10 or 1/2 of the total number of acorns. In the case of Auntie's method, Chloe's share is 2/5 of the total plus 10 (you might like to verify this). When the total is more than 100, 1/2 of the total is greater than 2/5 of the total plus 10.