The Improving Mathematics Education in Schools (TIMES) Project
The Improving Mathematics Education in Schools (TIMES) Project
Multiples, Factors and Powers
Number and Algebra : Module 19Years : 78
June 2011
Multiplication and division of whole numbers throw up many surprising things. This module encourages multiplicative thinking about numbers, and introduces ideas that are essential skills in fractions and algebra.
The ideas of this module are presented in purely arithmetical form, and no algebra is used except in some remarks that look forward to later work. The only numbers in the module are whole numbers, apart from the final paragraphs, where fractions are used so that the fifth index law can be presented in a more satisfactory form.
Students first meet the distinction between odd numbers and even numbers in early primary school, but it is useful everywhere in mathematics. Even numbers are multiples of 2, and more generally, multiples arise throughout mathematics and everyday life. The mass of a stack of bricks is a multiple of the mass of one brick. The number of pages in a packet of notebooks is a multiple of the number of pages in one notebook.
The factors of a number can be displayed using rectangular arrays. Some numbers, such as 30, can arise in many different ways as a product,
30 = 1 × 30 = 2 × 15 = 3 × 10 = 5 × 6 = 2 × 3 × 5,
whereas a number such as 31 can only be written trivially as the product 31 = 1 × 31. This idea leads to the classification of numbers greater than 1 as either prime or composite, and to a listing of all the factors of a number.
There are several groups of wellknown divisibility tests that can check whether a number is a factor without actually performing the division. These tests greatly simplify the listing of factors of numbers.
Repeated addition leads to multiplication. Repeated multiplication in turn leads to powers, and manipulating powers in turn relies on five index laws. Powers are introduced in this module, together with four of the five index laws.
We are used to comparing numbers in terms of their size. The highest common factor (HCF) and lowest common multiple (LCM) allow us to compare numbers in terms of their factors and multiples. For example, when we look at 30 and 12, we see that they are both multiples of 6, and that 6 is the greatest factor common to both numbers. We also see that 60 is a multiple of both numbers, and that 60 is the lowest common multiple of them (apart from 0). The HCF and LCM are essential for fractions and later for algebra.
Here is the usual definition of odd and even whole numbers.
Thus 10 is even and 11 is odd. We can demonstrate this by writing
10 = 5 + 5 and 11 = 5 + 5 + 1,
and we can illustrate this using arrays with two rows.
10 = 5 + 5 
11 = 5 + 5 + 1 
The array representing the even number 10 has the dots divided evenly into two equal rows of 5, but the array representing the odd number 11 has an extra odd dot left over.
When we write out the whole numbers in order,
the even and odd numbers alternate, starting with 0, which is an even number because
0 + 0 = 0.
This pattern occurs in all sorts of common situations:
Indeed, our concept of the number 2 is so different from our conceptions of all other numbers that we even use different language. We divide a pie between two people, but among three people. We identify two alternatives, but three options. The word ‘doubt’ is related to the Latin ‘duo’, the word ‘twofaced’ means ‘liar’, and the traditional number of the devil is 2.
EXERCISE 1
Students often come up with other possible definitions of even whole numbers:
What particular property of odd and even numbers does each illustrate?
Adding and subtracting odd and even numbers
There are several obvious facts about calculations with odd and even numbers that are very useful as an automatic check of calculations. First, addition and subtraction:
Proofs by arrays usually convince students more than algebraic proofs. The diagram below illustrates ‘odd plus odd equals even’, and shows how everything depends on the odd dot left over. The other cases are very similar.
+ =
EXERCISE 2
Draw four diagrams to illustrate the four cases of subtraction of odd and even numbers.
Multiplication of odd and even numbers
When we multiply odd and even numbers,
Proofs by arrays can be used here, but they are unwieldy. Instead, we will use the previous results for adding odd and even numbers. Here are examples of the three cases:
6 × 4 = 24, 5 × 4 = 20, 7 × 3 = 21.
The first and second products are even because each can be written as the sum
of even numbers:
6 × 4 = 4 + 4 + 4 + 4 + 4 + 4 and 5 × 4 = 4 + 4 + 4 + 4 + 4.
The third product can be written as the sum of pairs of odd numbers, plus an extra
odd number:
7 × 3 = (3 + 3) + (3 + 3) + (3 + 3) + 3.
Each bracket is even, because it is the sum of two odd numbers, so the whole sum is odd.
EXERCISE 3
What can we say about the quotients of odd and even numbers? Assuming in each case that the division has no remainder, complete each sentence below, if possible. Justify your answers by examples.
Using algebra for odd and even numbers
The previous results on the arithmetic of odd and even numbers can be obtained later after pronumerals have been introduced, and expansions of brackets and taking out a common factor dealt with. The important first step is:
EXERCISE 4
Obtain the previous results on the addition and multiplication of odd and even numbers by algebra. Begin, ‘Let the even numbers be 2a and 2b, and the odd numbers be 2a + 1 and 2b + 1, where a and b are whole numbers.’
Representing numbers by arrays
In the previous section, we represented even numbers by arrays with two equal rows, and odd numbers by arrays with two rows in which one row has one more dot than the other.
Representing numbers by arrays is an excellent way to illustrate some of their properties. For example, the arrays below illustrate significant properties of the numbers 10, 9, 8 and 7.
The number 10
There are two rectangular arrays for 10:
10 = 2 × 5 
10 = 1 × 10 
The first array shows that 10 can be factored as 10 = 5 × 2, which means that 10 is an even number.
The second array is trivial − every number can be factored as a product of itself and 1.
The convention used in these modules is that the first factor represents the number of rows, and the second factor represents the number of columns. The opposite convention, however, would be equally acceptable.
The number 9
The number 9 also has two rectangular arrays:
9 = 3 × 3 
9 = 1 × 9 
The first array shows that 9 is a square because it can be represented as a square array. The corresponding factoring is 9 = 3 × 3.
Because there is no 2row array, the number 9 is odd. The second array is the trivial array.
The number 8
The number 8 has two rectangular arrays and a threedimensional array:
8 = 2 × 4 
8 = 1 × 8 
8 = 2 × 2 × 2 
The third array shows that 8 is a cube because it can be represented as a cubic array.
The corresponding factoring is 8 = 2 × 2 × 2.
The first array shows that 8 is even, and the second array is the trivial array.
The number 7
The interesting thing about the number 7 is that it only has the trivial array, because 7 dots cannot be arranged in any rectangular array apart from a trivial array.
7 = 1 × 7
Numbers greater than 1 with only a trivial rectangular array are called prime numbers. All other whole numbers greater than 1 are called composite numbers.
We shall discuss prime numbers in a great deal more detail in the later module, Primes and Prime Factorisation. The usual definition of a prime number expresses exactly the same thing in terms of factors:
Here are the only possible rectangular arrays for the first four prime numbers:
2 
3 
5 
7 
Exercise 5
EXERCISE 6
EXERCISE 7
Rectangular arrays are not the only way that numbers can usefully be represented by patterns of dots.
1 
1 + 2 = 3 
1 + 2 + 3 = 6 
1 + 2 + 3 + 4 = 10 
1 + 2 + 3 + 4 + 5 = 15 
Thus the first few triangular numbers are: 1, 3, 6, 10, 15, 21,…
EXERCISE 8
EXERCISE 9
In the diagram to the right, two copies of the fourth triangular number have been fitted together to make a rectangle.
Explain how to calculate from this diagram that the 4th triangular number is 10. Hence calculate the 100th triangular number.
Multiples, common multiples and the LCM
We can arrange the multiples of 6 in increasing order,
0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96,…
so that they form a simple pattern increasing by 6 at each step.
Because 6 is an even number, all its multiples are even. The multiples of an odd number such as 7, however, alternate even, odd, even, odd, even,… because we are adding the odd number 7 at each step.
The number zero is a multiple of every number. Because of this, the word ‘multiple’ is sometimes used in the sense of ‘nonzero multiple’, but we will always add the word ‘nonzero’ when 0 is excluded.
The multiples of zero are all zero. Every other whole number has infinitely many multiples. This phrase ‘infinitely many’ has a very precise meaning — however many multiples you write down, there is always another multiple that you haven’t written down.
We can illustrate the multiples of a number using arrays with three columns and an increasing numbers of rows. Here are the first few multiples of 3:
0 
3 
6 
9 
12 
15 
18 
Rows and columns can be exchanged. Thus the multiples of 3 could also be illustrated using arrays with three rows and an increasing numbers of columns.
Division
The repeating pattern of common multiples is a great help in understanding division.
Here again are the multiples of 6,
0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96,…
If we divide any of these multiples by 6, we get a quotient with remainder zero.
For example,
24 ÷ 6 = 4 and 30 ÷ 6 = 5.
To divide any other number such as 29 by 6, we first locate 29 between two multiples of 6. Thus we locate 29 between 24 and 30. Because 29 = 24 + 5, we write
29 ÷ 6 = 4 remainder 5 or alternatively 29 = 6 × 4 + 5.
Notice that the remainder is always a whole number less than 6, because the multiples of 6 step up by 6 each time. Hence with division by 6 there are only 6 possible remainders,
0, 1, 2, 3, 4, 5.
This result takes a very simple form when we divide by 2, because the only possible remainders are 0 and 1.
In later years, when students have become far more confident with algebra, these remarks about division can be written down very precisely in what is called the division algorithm.
For example, we saw that 29 ÷ 6 = 4 remainder 5 means that
29 = 6 × 4 + 5 and 0 £ 5 < 6.
EXERCISE 10
The table above shows all the whole numbers written out systematically in 7 columns. Suppose that each number in the table is divided by 7 to produced a quotient and
a remainder.
EXERCISE 11
When 22 and 41 are divided by 6, their remainders are 4 and 5 respectively. Yet when their sum 63 is divided by 6, the remainder is 3, and is not 4 + 5 = 9. Explain.
Common multiples and the LCM
An important way to compare two numbers is to compare their lists of multiples. Let us write out the first few multiples of 4, and the first few multiples of 6, and compare the two lists.
The numbers that occur on both lists have been circled, and are called common multiples.
The common multiples of 6 and 8 are 0, 12, 24, 36, 48,…
Apart from zero, which is a common multiple of any two numbers, the lowest common multiple of 4 and 6 is 12.
These same procedures can be done with any set of two or more nonzero whole numbers.
EXAMPLE
Solution
Two or more nonzero numbers always have a common multiple — just multiply the numbers together. But the product of the numbers is not necessarily their lowest common multiple. For example, in the case above of 4 and 6, one common multiple is 4 × 6 = 24, but their lowest common multiple is 12.
EXAMPLE
Solution
The common multiples are the multiples of the LCM
You will have noticed that the list of common multiples of 4 and 6 is actually a list of multiples of their LCM 12. Similarly, the list of common multiples of 12 and 16 is a list of the multiples of their LCM 48.
This is a general result, which in Year 7 is best demonstrated by examples. In an exercise at the end of the module, Primes and Prime Factorisation, however, we have indicated how to prove the result using prime factorisation.
We will have more to say about the LCM once the HCF has been introduced.
Factors, common factors and the HCF
This can be restated in terms of the multiples of the previous section:
The number zero is a multiple of every number, so every number is a factor of zero. On the other hand, zero is the only multiple of zero, so zero is a factor of no numbers except zero. These rather odd remarks are better left unsaid, unless students insist. They should certainly not become a distraction from the nonzero whole numbers that we want to discuss.
The product of two nonzero whole numbers is always greater than or equal to each factor in the product. Hence the factors of a nonzero number like 12 are all less than or equal to 12. Thus whereas a positive whole number has infinitely many multiples, it has only finitely many factors.
The long way to find all the factors of 12 is to test systematically all the whole numbers less than 12 to see whether or not they go into 12 without remainder. The list of factors of 12 is
1, 2, 3, 4, 6 and 12.
It is very easy to overlook factors by this method, however. A far more efficient way, is to look for pairs of factors whose product is 12. Begin by testing all the whole numbers 1, 2, … that could be the smaller of a pair of factors with product 12,
1, 2, 3,…
We stop at 3 because 42 = 16 is greater than 12, so 4 cannot be the smaller of a pair.
Now we add, in reverse order, the complementary factor of each pair, that is, 12 ÷ 3 = 4, 12 ÷ 2 = 6 and 12 ÷ 1 = 12,
We can display these pairs of factors by writing the 12 dots in all possible rectangular arrays:
1 × 12 
2 × 6 
3 × 4 
For a larger number such as 60, the method recommended here has the following steps:
Thus the factors of 60 are
EXERCISE 12
Use this method to write down all the factors of 72 and 160.
Common factors and the HCF
Another important way to compare two numbers is to compare their lists of factors.
Let us write out the lists of factors of 18 and 30, and compare the lists.
The numbers that occur on both lists have been circled, and are called the common factors.
The common factors of 18 and 30 are 1, 2, 3 and 6.
The highest common factor is 6.
As with common multiples, these procedures can be done with any list of two or more whole numbers.
The HCF is also known as the ‘greatest common divisor’, with corresponding initials GCD.
EXAMPLE
Solution
Any collection of whole numbers always has 1 as a common factor. The question is whether the numbers have common factors greater than 1.
Every whole number is a factor of 0, so the common factors of 0 and say 12 are just the factors of 12, and the HCF of 0 and 12 is 12. A nonzero whole number has only a finite number of factors, so it has a greatest factor. Two or more numbers always have a HCF because at least one of them is nonzero. These are distractions from the main ideas.
EXAMPLE
Solution
The common factors are the factors of the HCF
You will have noticed that the list of common factors of 18 and 30 is actually a list of factors of their HCF 6. Similarly, the list of common factors of 30 and 75 is a list of the factors of their HCF 15.
Again, this is a general result, which in Year 7 is best demonstrated by examples. An exercise at the end of the module Primes and Prime Factorisation indicates how to prove the result using prime factorisation.
Two relationships between the HCF and LCM
The two relationships below between the HCF and the LCM are again best illustrated by examples in Year 7, but an exercise in the module Primes and Prime Factorisation indicates how they can be proven.
The first relationship is extremely useful, and is used routinely when working with common denominators of fractions.
The converse is also true, although it does not arise so often.
For example,
The numbers 4 and 9 have HCF 1, and their LCM is their product 36.
The numbers 6 and 7 have HCF 1 and their LCM is their product 42.
EXAMPLE
Solution
The second relationship is not so obvious, and needs to be brought out by examples.
For example, the numbers 4 and 6 have HCF 2 and LCM 12, and 2 × 12 = 4 × 6.
EXAMPLE
Confirm this relationship for:
a10 and 15 b 12 and 9 c 40 and 10
Solution
The threepart example below indicates how this relationship can be proven from the relationship above, although such a proof would be unsuited for most Year 7 students.
EXERCISE 13
When we multiply a number by itself, we usually use a more concise notation,
3 = 31  
3 × 3 = 32  (read as ‘3 squared’)  
3 × 3 × 3 = 33  (read as ‘3 cubed’)  
3 × 3 × 3 × 3 = 34  (read as ‘3 to the power 4’ or as ‘3 to the 4th’)  
3 × 3 × 3 × 3 × 3 = 35  
3 × 3 × 3 × 3 × 3 × 3 = 36 
The terms ‘3 squared’ for 32 and ‘3 cubed’ for 33 come from geometry. As we saw earlier in the module, we can arrange 32 dots in a square and 33 dots in a cube:
32  33 
As mentioned before, there is no straightforward geometrical representation of
higher powers.
Exercise 14
Exercise 15
Exercise 16
Adding successive odd numbers gives successive squares:
1 = 1  
1 + 3 = 4  
1 + 3 + 5 = 9  
1 + 3 + 5 + 7 = 16  
1 + 3 + 5 + 7 + 9 = 25  
1 + 3 + 5 + 7 + 9 + 11 = 36  
… 
Dissect a square array of 25 dots to show why this is happening.
Exercise 17
[A rather difficult challenge activity] Adding successive cubes gives the squares of the triangular numbers:
13  = 1  = 12  
13 + 23  = 9  = 32  
13 + 23 + 33  = 36  = 62  
13 + 23 + 33 + 43  = 100  = 102  
13 + 23 + 33 + 43 + 53  = 225  = 152 
Successive powers of a number
Powers of numbers are used extensively later in the study of logarithms and of combinatorics. It is useful to be able to compute or remember some smaller powers quickly, and recognise them.
The powers of 2 are: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192,…
The powers of 3 are: 3, 9, 27, 81, 243, 729,…
The powers of 4 are every second power of 2.
The powers of 5 are: 5, 25, 125, 625, 3125,…
The powers of 6 are: 6, 36, 216,…
The powers of 7 are: 7, 49, 343,…
The powers of 8 are every third power of 2.
The powers of 9 are every second power of 3.
The powers of 10 are: 10, 100, 1000, 10 000, 100 000, 1 000 000,…
The powers of 16 are every fourth power of 2.
Our base 10 placevalue system displays every number as a sum of multiples of powers of 10.
Computers use a base 2 placevalue system, so computer programmers need to know the powers of 2, and must be able to convert numbers quickly to a sum of powers of 2.
Exercise 18
Write 201 as a sum of powers of 2. Hence write 201 in base 2 notation.
There are several straightforward tests for divisibility that are very useful when factoring numbers. They all have their origin in the base 10 that we use for our system of numerals.
Divisibility by powers of 2 and powers of 5
The tests for divisibility by powers of 2 and 5 arise from the fact that 10 = 2 × 5.
Because 10 is a multiple of 2, every multiple of 10 is a multiple of 2. Thus to test whether a number is divisible by 2, we only need to look at the last digit.
Similarly, 10 is a multiple of 5, so every multiple of 10 is a multiple of 5. Thus to test whether a number is divisible by 5, we only need to look at the last digit.
For example:
The number 864 ends with 4, so it is divisible by 2, but not by 5.
The number 1395 ends with 5, so it is divisible by 5, but not by 2.
The number 74 830 ends with 0, so it is divisible both by 2 and by 5.
The squares 22 = 4 and 52 = 25 are factors of 100, so every multiple of 100 is a multiple of 4 and of 25. Thus to test for divisibility by 4 and 25, we only need to look at the last two digits.
Similarly, 23 = 8 and 53 = 125 are factors of 1000, so
These tests can be continued for higher powers of 2 and 5.
Example
Solution
Divisibility by 9 and 3
The tests for divisibility by 9, and in turn by 3, arise from the fact that 9 is 1 less than 10.
When 1 or any power of 10 is divided by 9, the remainder is 1. For example,
1 ÷ 9  = 0 remainder 1  
10 ÷ 9  = 1 remainder 1  
100 ÷ 9  = 11 remainder 1  
1000 ÷ 9  = 111 remainder 1 
Multiplying each by a single digit number less than 9,
4 ÷ 9  = 0 remainder 4  
50 ÷ 9  = 5 remainder 5  
200 ÷ 9  = 22 remainder 2  
9000 ÷ 9  = 999 remainder 9; better written as 1000 remainder 0. 
The remainder when 9254 is divided by 9 is the same as the remainder when
9 + 2 + 5 + 4 = 20 is divided by 9. The general result is
Because 9 is a multiple of 3, the remainders after division by 3 follows a similar pattern,
4 and 4 have the same remainder after division by 3.
50 and 5 have the same remainder after division by 3.
200 and 2 have the same remainder after division by 3.
9000 and 9 have the same remainder after division by 3.
Adding these results, 9254 and 9 + 2 + 5 + 4 = 20 have the same remainder when divided by 3. The general result is
For example,
167 ÷ 9 = 18 remainder 5  and  167 ÷ 3 = 55 remainder 2  
(1 + 6 + 7) ÷ 9 = 1 remainder 5  (1 + 6 + 7) ÷ 3 = 4 remainder 2 
Example
Find the remainders when 62 574 is divided by 9 and by 3.
Solution
The sum of the digits is 6 + 2 + 5 + 7 + 4 = 24, and 2 + 4 = 6.
Hence 62 574 has remainder 6 when divided by 9, and remainder 0 when divided by 3.
With most students in Years 7−8, however, only the corresponding divisibility tests
are appropriate:
Example
Test for divisibility by 3 and by 9:
a 71 325 618 b 36 867 924
Solution
Divisibility by 11
The test for divisibility by 11 arises from the fact that 11 is 1 more than 10, but 9 × 11 = 99 is 1 less than 100.
When 1 or any power of 10 is divided by 11, the remainder is alternately 1 and 10, but we shall regard the remainder 10 as being remainder −1.
1 ÷ 11  = 0 remainder 1  
10 ÷ 11  = 0 remainder 10; which we will regard as remainder −1  
100 ÷ 11  = 9 remainder 1  
1000 ÷ 11  = 91 remainder 10; which we will regard as remainder −1  
10 000 ÷ 11  = 909 remainder 1  
100 000 ÷ 11  = 9091 remainder 10; which we will regard as remainder −1 
Multiplying each result by a single digit number,
4 ÷ 11  leaves remainder 4  
20 ÷ 11  leaves remainder −5  
500 ÷ 11  leaves remainder 2  
3000 ÷ 11  leaves remainder −9  
90 000 ÷ 11  leaves remainder 3  
800 000 ÷ 11  leaves remainder −8 
Adding these results, the remainder when 893 524 is divided by 11 is the same as the remainder when 4 − 2 + 5 − 3 + 9 − 8 = 7 is divided by 9. It was important here to write the alternating sum of the digits starting at the righthand end with the units. The general result is,
For example,
91 627 ÷ 11 = 8329 remainder 8 and (7 − 2 + 6 − 1 + 9) ÷ 11 = 1 remainder 8.
Example
Find the remainders after division by 11 of:
a 62 578 b 937 565
Solution
As with 3 and 9, only the corresponding divisibility test is usually appropriate in Years 7−8:
Example
Test for divisibility by 11:
a 71 325 618 b 71 938 471
Solution
Divisibility by numbers that are not prime powers
We can combine the previous tests for divisibility by testing separately for divisibility by the highest power of each prime. Here are examples of some such tests:
Divisibility by powers of 10 is particularly simple—just count the number of zeroes.
Example
Solution
The multiplication table is one of the bestknown objects in arithmetic. It is formed by doing nothing more that writing out the first twelve nonzero multiples of each number from 1 to 12 in twelve successive rows. (Or perhaps they were written out in twelve successive columns.)
Despite the simplicity of its construction, it is a very powerful object indeed, and well justifies the recommendation to learn it by heart. Here are some of its many properties — students routinely find many more.
Students often see many more patterns in this table. The following exercise gives some less obvious properties, but the proofs are omitted, because they require quite serious algebra. Once sequences and series are being studied, perhaps in Year 11, the table is well worth revisiting because of the insights it can give into sequences and into the use of algebra.
Exercise 19
There are five useful laws, collectively called the index laws, that help us manipulate powers. At this stage, they are best established by examples and learnt verbally.
After algebra has been introduced, however, they can be presented in their more
familiar algebraic formulation. For later reference, they are summarised here verbally
and algebraically:
Multiplying powers with the same base
If we write out the powers as continued products, we can quickly see what is happening when we multiply powers with the same base.
75 × 73 = (7 × 7 × 7 × 7 × 7) × (7 × 7 × 7) = 78
There are 5 + 3 = 8 factors of 7 in the product, so the product is 78.
Dividing powers with the same base
We can use the same approach with dividing powers with the same base.
75 ÷ 73 = (7 × 7 × 7 × 7 × ) ÷ (7 × 7 × 7) = 72
There are 5 − 3 = 2 factors of 7 in the quotient, so the quotient is 72.
EXAMPLE
Simplify each expression, leaving the answer as a power.
a 32 × 35 b 57 ÷ 53 c 109 ÷ 103 d 129 × 125 ÷ 122
Solution
a 32 × 35 = 37 b 57 ÷ 53 = 54 c 109 ÷ 103 = 106 d 129 × 125 ÷ 122 = 1212
The examples above have carefully avoided questions like 73 ÷ 75, where the divisor is a higher power than the dividend. Such questions are better written as fractions,
= or = 72
The first form can be covered once fractions have been introduced. The second form requires negative indices, which are usually regarded as a little too sophisticated for Years 7−8.
Raising a power to a power
We often need to deal with expressions such as (73)4 in which a power is raised to a power. Writing out just the outer power allows us to apply the previous law.
(73)4  = 73 × 73 × 73 × 73  
= 712 
There are 3 × 4 factors of 7 in the expression, so it simplifies to 712.
(73)4 = 712
EXAMPLE
Simplify each expression, leaving the answer as a power:
a (65)7 b (103)5 c (42)3 × (43)2 d (115)4 ÷ (116)3
Solution
a 
(65)7 = 635 
b 
(103)5 = 1015 

c 
(42)3 × (43)2 = 46 × 46 = 412 
d 
(115)4 ÷ (116)3 = 1120 ÷ 1118 = 112 
The power of a product
The fourth law deals with a power of a product. Again, we can write out the power.
(2 × 3)4  = (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3)  
= (2 × 2 × 2 × 2) × (3 × 3 × 3 × 3)  
= 24 × 34 
These steps rely heavily on the anyorder property of multiplication to regroup the
4 factors of 2 and the 4 factors of 3.
(2 × 3)4 = 24 × 34
Example
Write each expression without brackets, leaving the answer in index notation.
a (3 × 7)4 b (8 × 12)6 c (3 × 5)4 × (3 × 7)3 d (5 × 7 × 9)6 ÷ (55)2
Solution
a  (3 × 7)4 = 34 × 74  b  (8 × 12)6 = 86 × 126  
c  (3 × 5)4 × (3 × 7)3  = 34 × 54 × 33 × 73  d  (5 × 7 × 9)16 ÷ (55)2  = 516 × 716 × 916 ÷ 510  
= 37 × 54 × 73  = 56 × 716 × 916 
The power of a quotient
The final index law deals with the power of a quotient. It is clumsy to formulate this law using the division sign, so fraction notation for division is used for the only time in this module.
= × × ×  
= 
The same process can be done with the power of any fraction, giving the general result,
=
EXAMPLE
Simplify each expression, evaluating the resulting powers:
a b c d
Solution
a 
4 = = 
b 
= = 

c 
= = 
d 
= = = = 5 
The index laws and mental arithmetic
‘A power of a product is the product of powers’ is useful in mental arithmetic, both forwards and in reverse. For example,
204  = (2 × 10)4  32 × 125  = 25 × 53  
= 24 × 104  = 22 × (2 × 5)3  
= 16 × 10 000  and  = 22 × 103  
= 160 000  = 4 × 1000  
= 4000 
Because our system of numerals has base 10, isolating powers of 10 is always good practice. In the first example, we factored 20 as 20 = 2 × 10. In the second example, we combined factors of 2 and factors of 5 to make factors of 10.
Exercise 20
Show how to calculate these expressions using the index laws to work with powers of 10.
a 30004 b 4 000 0003 c 625 × 8000 d 40003 × 503
The HCF and LCM are essential for the study of fractions, which are introduced in the module Fractions. To express a fraction in simplest form, we cancel the HCF of the numerator and denominator, and to add and subtract fractions, we usually use the LCM of their denominators as a common denominator,
= and + = + = .
The HCF of two algebraic expressions is important in factoring. The first step in the systematic factoring of an expression is to take out the HCF of the terms:
3ax2 − 6a2x = 3ax(x − 2a).
Factoring by taking out the HCF is first considered in the module Algebraic Expressions, and developed further in the module, Negatives and the Index Laws.
Prime numbers are the building blocks of all whole numbers when we take them apart by factoring. For example, we can write 60 as a product of primes,
60 = 22 × 3 × 5.
Every composite number has one and only one prime factorisation. Prime factorisation is the central concern of the module Primes and Prime Factorisation. This module also shows how prime factorisation yields more systematic approaches to the calculations of the HCF and LCM.
The five index laws have been introduced in the present module in the context of whole numbers, although it was mentioned that the fifth law, and a clearer statement of the second law, require fractions. The modules, Negatives and the Index Laws and Special Expansions and Algebraic Fractions extend them to rational numbers and also give them an algebraic formulation, although the indices are still only nonzero whole numbers. The later module The Index Laws extends the indices to any rational number.
The presentation in this module has been carried out arithmetically, and entirely within the whole numbers. There have been no algebra or fractions except in the occasional remark and exercise that looks forward to later content. Arrays, rather than areas, have been used to represent products, so that the wholenumber aspect of the ideas can be emphasised. It is important in later years, when algebra has been used to generalise the content, not to lose the arithmetic intuition of the present module.
All the present material was expounded by the ancient Greek mathematicians. The idea of representing numbers by arrays was developed particularly by Pythagoras and his school, who developed theories about what we now call ‘figurate numbers’. These include the triangular numbers discussed in this module, and pentagonal and tetrahedral numbers.
Exercise 2
Here is ‘odd minus odd equals even’. The other cases are similar.
− =
Exercise 3
Exercise 4
For addition,
2a + 2b  = 2(a + b)  
(2a + 1) + 2b  = 2(a + b) + 1  
(2a + 1) + (2b + 1)  = 2(a + b + 1) 
For multiplication,
(2a) × (2b)  = 2 × 2ab  
(2a) × (2b + 1)  = 2 × a(2b + 1)  
(2a + 1) × (2b + 1)  = 4ab + 2a + 2b + 1  
= 2(2ab + a + b) + 1 
Exercise 5
Exercise 6
Exercise 7
Exercise 8
Exercise 9
The diagram shows that the 4th triangular number is (4 × 5) ÷ 2 = 10.
Similarly, the 100th triangular number is (100 × 101) ÷ 2 = 5050.
Exercise 10
Exercise 11
The remainders when 22 and 41 are divided by 6 are 4 and 5, whose sum is 9.
But 9 = 6 + 3, so the remainder when dividing 63 by 6 is 3.
Exercise 12
The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
The factors of 160 are 1, 2, 4, 5, 8, 10, 16, 20, 32 40, 80, 160.
Exercise 13
(Notice that the resulting numbers in parts b and c are the same.)
Exercise 14
Exercise 15
Exercise 16
Exercise 18
201 = 128 + 64 + 8 + 1 = 27 + 26 + 23 + 1, so in base 2 notation, 201 = 11 001 001.
Exercise 19
Exercise 20
a  30004  = 34 × (103)4  
= 81 × 1012  
= 81 000 000 000 000 
b  4 000 0003  = 43 × (106)3  
= 64 × 1018  
= 64 000 000 000 000 000 000 
c  625 × 8000  = 54 × 23 × 103  
= 5 × 103 × 103  
= 5 000 000 
d  40003 × 503  = (4000 × 50)3 

= 200 0003 

= 8 000 000 000 000 000 
The Improving Mathematics Education in Schools (TIMES) Project 20092011 was funded by the Australian Government Department of Education, Employment and Workplace Relations.
The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations.
© The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICEEM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons AttributionNonCommercialNoDerivs 3.0 Unported License.
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