I was delighted to see, in Equals Vol. 5 No.2, p. 9, a set of seven possible answers to a division problem in which the differences in answer depended on the context of the problem. As the following explanations show, the context is crucial. The problem, you may remember, was sharing 23 equally between 6. The analysis below may well give those who ?can do? mathematics some understanding of those who ?can?t?. It may be connected to the difficulty in abstracting the `correct' answer of 3. 833333..., or 3 5/6, from the numerical problem of 23 shared equally between 6.1.

Example 1 was: 23 oranges shared equally between 6 people How many oranges does each person get? The ?answer? given was 3 oranges and 5/6 of an orange.
However, how do you divide an orange exactly into sixths? I haven?t found one, yet, with six segments; 8, 9, and 10 segments, yes, but not 6. (and even then the segments would have to be of equal size or mass). You could, perhaps, divide the 5 remaining oranges by weight, but then some people involved might get a higher proportion of peel than others, and hence it would not be an equal distribution. Those involved might agree to squeeze all 23 oranges and share the resultant juice and pulp by volume and mass; but I doubt that this scenario was what Mr. Vertes was thinking of at the time. Within this mathematical solution we have also assumed that the oranges were all the same size, shape, and texture. Is this reasonable? Do we always take the first 3 oranges on display, or does the one at the back look sweeter?

Example 2: 23 people go in cars. Each car can take at most 6 people. How many cars are needed for the group?
The simplistic answer is 4, yet we have assumed that the people are about the same size. However, if 2 or more of them are over 130 kgs, the answer might have to be 5 or more. We have also assumed no personality or creed difficulties. In some areas of the world a woman would not sit next to a man, or a person of one faith might refuse to sit with one or more of another faith.

Example 3: 23 paper clips are shared amongst 6 people. How are the paper clips shared?
The answer 3 remainder 5 is obviously ?correct?. We probably have identical paper clips, and we have 5 spares. But why should such a problem as this arise in the first place? Perhaps the six people have been set a task which involves the use of 3 paper clips each? and the other 5 are also needed by the group of 6? Are we to make sure that they all receive them at the same time? and, if not, we will, to some degree, not have shared them out equally. (Possible task for the 6 people each person makes a 3 link chain. When all 6 have been made they join them into one long chain using the remaining 5.) But then the question should have been, ?How many paper clips do you need to...??.

Examples 4 and 5: those of sharing winnings and sharing debts, are very good examples of real life mathematics, showing us how ?correct? answers can be manipulated by those who have the money to begin with. In the case of sharing winnings of £23 equally between 6 people, they each gain £3.83. However 6 lots of £3.83 amount to £22.98, and so they have a collective loss of 2 pence. When 6 people have to share a bill of £23 pounds, they are each charged £3.84. Again they suffer a collective loss, this time of 4 pence, as 6 lots of £3.84 is £23. 04. It is true that the people involved could makes arrangements amongst themselves if they shared the same bills on a regular basis. If two of them agreed to paid £3.84 on the first occasion whilst the others paid £3.83 pounds, and the ?pair? who paid the larger amount was rotated, we could have created a way of sharing the bill ?equally?.

However, it is unlikely, to say the least, that we could arrange a similar set of circumstances to assist us in sharing the winnings ?equally?, as winnings do not fall into the category of regular events.

Example 6: 23 full books of stamps, and an offer that allows 6 books to be exchanged for a CD. How many CDs would you get?
Here, obviously, you would get 3 CDs ( providing you arranged that the exchange took place before the closing date specified in the offer, that they had not already run out of those CDs on offer ? and that the offer was not a bogus one); but what would you do with the other 5?

    a) You could contact the company involved , and hope that they would accept 5 instead of 6 for the 4th CD.
    b) You could ask them to put you in touch with others who also had `spares'.
    c) You could contact a radio station and ask them to publish your dilemma.
Or,
    d) as I suspect would happen in many, if not most, cases, you could throw them away.

In these circumstances the equality of the sharing depends on the astuteness and perseverance of the people to whom the offer might be made.

Example 7: 23 is divided by 6, using a calculator. What is the answer?
Here, providing the calculator is used correctly, the answer is guaranteed, and this will, I?m sure make the Pure Mathematicians happy. However, as it can?t give the correct answer in the other 6 practical questions, how accurate, or indeed useful, is it?

Happy mathematical thinking!