Fair Shares

This activity deals with partitive division, that is, division where the number of parts is known but the amount of each share is not. Students learn that they can turn a complex division problem into a simpler one by simplifying both the dividend (the starting number) and the divisor (the number doing the dividing), using a common factor.
You could introduce the mathematics of question 1 with a simple example like the one below to illustrate the number properties. By doing this, you move the focus from the answer, which the students already know or can work out easily, to the number properties involved. By keeping the numbers manageable, you can use materials to demonstrate the transformations on the quantities involved.
Suppose that there are 32 biscuits to be shared among 8 people. How can the numbers in the problem be altered without affecting the size of the share?

Halving both the number of biscuits and the number of people who share them keeps the size of the shares the same: 32 ÷ 8 = 4, 16 ÷ 4 = 4, and 8 ÷ 2 = 4. The same is true no matter what number the numbers of biscuits and people are divided by. The size of the shares is also unaffected if the numbers of biscuits and people are doubled, trebled, or multiplied by any number at all, although this operation will seldom make a division problem easier. Algebraically, the principle can be expressed in this way: if a ÷ b = c, then a/n ÷ b/n = c where n is any number except zero.
Students will be able to use this method as long as they can spot a value for n that works, that is, a value that is a common factor of both numbers. For example, in 198 ÷ 6 =, students won’t be able to reduce the problem to 99 ÷ 3 = unless they realise that both 198 and 6 are even numbers and so have a common factor of 2.
Sometimes, it will take too much work to simplify a division problem using this method and another strategy is more suitable. Students need to recognise when halving, thirding, and so on of both numbers is an effective strategy and when it is not. Effectiveness depends on how easy it is to find a common factor and how easy it is to divide by that common factor. Take, for example, 657 ÷ 9 =. Both 657 and 9 are divisible by 3, but the difficulty of dividing 657 by 3 means that, for many students, standard place value methods are likely to be a better choice. For example: 630 ÷ 9 = 70; 27 ÷ 9 = 3; 70 + 3 = 73.
In question 1, the unknown is the total number of biscuits on the table; in question 2, it is the size of each share. It is important that students meet division problems posed in different ways like this so that they come to understand the principle of inverse or reversibility.
Students need to be able to process question 3 without first solving the problem posed at the beginning: 168 ÷ 24 =. That is, they need to work backwards through a number of equivalent statements:

 In this way, they operate on statements of equality while accepting the lack of closure that comes from not knowing the quotient (answer). They know that must be half of 168 because 12 is half of 24, and so on. This is the same kind of thinking that underpins equivalent fractions, for example, 
The problems in question 4 can be solved in similar ways, but because the numbers are deliberately larger, the students will have to find the factors of the divisors (36, 28, and 27) before they can work backwards.
For question 5, the students should be very sure that their equations can be sensibly solved by strategies such as halving and doubling, place value, or working backwards, before they try them out on a classmate.
 

Answers to Activity

1. The completed equations are:
20 ÷ 4 = 5,
so 40 ÷ 8 = 5,
so 60 ÷ 12 = 5,
so 30 ÷ 6 = 5,
so 15 ÷ 3 = 5.
The pattern in these division equations is that doubling or halving both numbers produces the same answer. (60 ÷ 12 = 5, from 20 ÷ 4 = 5, is trebling.)
2. a. Grace used halving of both numbers to go directly from 72 ÷ 18 = 4 to 36 ÷ 9 = 4. Some other divisions using doubling and trebling patterns and basic multiplication
facts are: 12 ÷ 3 = 4, 24 ÷ 6 = 4, 48 ÷ 12 = 4, and 144 ÷ 36 = 4.
b. They will all have the same answer because both numbers in each division statement are either doubles or halves of other division statements that have an answer of 4. Sharing half as many biscuits among half as many people gives the same share.
3. 168 ÷ 24 = 84 ÷ 12 = 42 ÷ 6 = 7 biscuits per person;
so 12 x 7 = 84 (at the 12-seat table)
and 6 x 7 = 42 (at the 6-seat table);
4 x 7 = 28 (at the 4-seat table),
and 3 x 7 = 21 (at the 3-seat table); or 42 ÷ 6 = 7, so 84 ÷ 12 = 7, 21 ÷ 3 = 7,
and 28 ÷ 4 = 7.
4. a. 11 biscuits. (396 ÷ 36 = 132 ÷ 12 = 66 ÷ 6 = 11)
b. 8 biscuits. (224 ÷ 28 = 112 ÷ 14 = 56 ÷ 7 = 8)
c. 9 biscuits. (243 ÷ 27 = 81 ÷ 9 = 9)
5. Problems will vary