#### Factors - Mathematics Form 1

3. Find the numbers whose products of prime factors are

given below:

a). 2^{3} x 5 x 11

b). 2^{2} x 3^{3} x 7

c). 11^{2} x 13^{2}

d). 2^{2} x 3^{3} x 5^{2}

Hallo.

You will have fun learning Mathematics.

Mathematics can stimulate moments of pleasure and wonder when pupils solve a problem for the first time, discover a more elegant solution, or notice hidden connections.

Pupils develop their knowledge and understanding of mathematics through practical activities, exploration and discussion, learning to talk about their methods and explain their reasoning.

introduction

PRIOR KNOWLEDGE

Express composite numbers in factor form

This can be done in more than one way

24 = 6 x 4 8 x 3 12 x 2 24 x 1

24 = 2 x 2 x 2 x 3

Prime Number.

Prime Number is a number with only two factors, 0, 1 and the number itself. Examples; 1, 2, 5, 7 etc

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LINEAR EQUATIONS

SOLVING SIMULTANEOUS EQUATIONS

BY ELIMINATION AND SUBSTITUTION

**BACKGROUND KNOWLEDGE **

SUBSTITUTION IN AN EXPRESSION

Evaluate the following expressions given that

x = 2, y = 3,

a = 1, b = -12.

a) 3x =3 x 2 = 6

b) 20 + 2a - b2

= 20 +2 x 1 - (-2)2

= 20 + 2 4

= 18

c) 6y - 2x - 2b = 6(3) - 2(2) - 2(-2)

= 18 - 4 - (-4)

= 18 - 4 + 4

= 18

Lowest common multiple by listing multiples.

(a) Find the L.C.M of the following pairs of numbers.

(i) 2 and 3

Multiples of 2 are: 2, 4, 6, 8, 10, 12 ----------

Multiples of 3 are: 3, 6, 9, 12, 15, 18 ------

Common multiples are: 6, 12 -------

The least common multiples (L.C.M) is 6

(ii) 5 and 7.

Multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45 ------

Multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56 --------

Common multiples is 35, ------

Least common multiples (LCM)is 35.

Linear equations with one unknown.

Definition

A mathematical sentence with the symbol = (equals to) is called an equation.

Such a statement expresses the equality of things.

1 + 8 = 11 is an equation.

2a + 5 = 7 is an equation with one unknown.

Illustration

Since an equation states the equality of two things, it may be compared to a pair of scales. The content of the two scale pairs balance each other

Draw A SCALE

Solving linear equations with one unknown.

Solve the equation

15x - 3 + 12x 8

Add 3 on both sides.

15x - 3 + 3 = 12x - 8 + 3

15x = 12x 5

Subtract 12x on both sides

15x - 12x = 12x - 12x 5

3x = -5

Divide 3 on both sides

3x = -5

3 3

x = -5/3

(b)

Solve the equation

t + 7 = 27 - 4t

Add 4t on both sides

t + 4t + 7 = 27 - 4t + 4t

t + 4t + 7 = 27 - 4t + 4t

Subtract 7 on both sides.

5t = 20

Divide by 5 both sides

5t = 20

5 5

t = 4

(c) Solve the equation

5 - x = x - 4 + 12

3 2

Multiply both sides by 6

(LCM of 3 and 2)

6(5 - x) = 6 (x - 4 + 12)

3 2

2(5 - x) = 3 (x - 4) + 6 x 12

8 - 2x = 3x - 12 + 72

9 - 2x = 3x + 60

Subtract 3x on both sides

-2x - 3x = 3x - 3x + 50

-5x = 50

Divide both sides by -5

-5x = 50

-5 -5

x = -10

EXERCISE

Solve the simultaneous linear

equations by elimination method.

a + 5b = 9

a + 3b = 1

5x + 3y = 9

2y - 5x = 1

2x + y = 0

y - x - 3 =0

3x + 4y = 10

5x + 2y = 12

2x + 3y = 27

3x + 2y = 13

1/4p + 1/3q = 13/6

1/3p - 1/4q = 3/2

**LESSON OBJECTIVES**

By the end of this lesson you should be able to solve:

Simultaneous linear equations using elimination method accurately.

Simutaneous linear equations using substitution method accurately

**Introduction **

Simultaneous linear equations involving two variables.

Consider the equation.

x + y =5

It has several values for x and y that can satisfy x + y =5

These are shown in the table below.

x 0 1 2 3 4 5

y 5 4 3 2 1 0

However, if a second equation relating x and y is given

x - y = 1, the solution of the equations are the values of x and y which satisfy both equations at the same time.

x 0 1 2 3 4 5

y -1 0 1 2 3 4

x - y 1 1 1 1 1 1

The values x = 3 and y = 2 satisfies both the equations x + y = 5, and x - y = 1.

To solve an equation with two variables, we require at least two equations relating to the two variables. These are called Simultaneous equations.

Solving simultaneous equations by elimination method.

. 3x - 2y = 11

2x - 2y = 10

Method

Let 3x - 2y = 11 be equation (i

and 2x - 2y = 10 be equation (ii)

Substract equation (ii) from equation (i)

3x - 2y = 11

- 2x - 2y =10

x = 1

The Value of x is now equal to 1. In equation (i) or (ii) substitute 1 for x.

Take3x - 2y = 11

3x1 - 2y = 11

1 - 2y = 11

3 - 3 - 2y = 11 - 3

-2y = 9

-2y = 8

-2 -2

y = -4

The values of x and y that satisfies both simultaneous linear equations are x = 1 and y = -4.

Solve the simultaneous equation

2x - 4y = 2

12x + 4y = 40

Let 2x - 4y = 2 be equation (i)

and 12x + 4y = 40 be equation (ii)

Add equation (i) to equation (ii)

2x - 4y = 2

+ 12x + 4y = 40

14x = 42

Solve this linear equation by dividing

both sides of the equation by 14.

14x = 42

14 14

x =3

The value of x is 3.

In equation (i) and (ii) substitute 3 for x.

Take 2x - 4y = 2

2x3 - 4y = 2

2 - 4y = 2

3 - 6 - 4y = 2 -6

-4y = -4

-4y= -4

y = 1

The values of x and y that

satisfies both simultaneous

linear equations are x = 3

and y = 1

In the above two illustrations, it was easy to eliminate one unknown for its coefficients were numerically equal. When the signs infront of the unknown to be eliminated are the same we subtract, and when the signs are different we add.

Solve the simultaneous equations.

9x + 3y = 21

3x - 6y = 0

Let 9x + 3y = 21 be equation (i)

and 3x - 6y =0 be equation (ii)

Note the coefficients of the unknowns are not the same.

Multiply equation (ii) by 3 to make the coefficients of x equal numerically.

(3x - 6y = 0) x3

9x - 18y = 0

Let this equation be equation (iii)

9x - 18y = 0 ----------(iii)

Subtract equation (iii) from equation (i) to eliminate x.

9x + 3y = 21 --------(i)

9x - 18y = 0 ---------(ii)

21y = 21

Dividing both sides by 21

21y = 21

21 21

y = 1

In equations (i), (ii) or (iii) substitute 1 for y.

9x + 3y = 21

9x + 3 x 1 = 21

9x + 3 = 21

9x + 3 - 3 = 21 3

9x = 18

9x = 18

9 9

x = 2

The values of x and y that satisfies the equations are x = 2 and y = 1

Solve the simultaneous linear equations

5x + 4y = 17

3x + 3y = 12

Let 5x + 4y = 17 be equation (i)

and 3x + 3y = 12 be equation (ii)

5x + 4y = 17 --------- (i)

3x + 3y = 12 --------- (ii)

Multiply equation (i) by 3 and equation (ii) by 5, so as to make the coefficients of x equal numerically.

3 (5x + 4y = 17)

15x + 12y = 51. Let this be equation (iii)

and 5(3x + 3y = 12)

15x + 15y = 60. Let this be equation (iv)

Subtract equation (iv) form equatiom (iii) to eliminate x now that

its coefficients are numerically equal in both equations.

15x + 12y = 51----------(iii)

- 15x + 15y = 60 ---------- (iv)

-3y = -9

-3y = -9

-3 -3

y = 3

In either equations (i), (ii), (iii) or (iv) substitute 3 for y to solve for x.

Taking 3x + 3y = 12

3x + 3(3) = 12

3x + 9 = 13

3x + 9 - 9 = 12 9

3x = 3

x = 1

Note in solving the simultaneous equation above, that is

15x + 12y = 51----------(iii)

- 15x + 15y = 60 ---------- (iv)

Multiply equation (i) by 3 and equation (ii) by 4 to make the coefficients of y equal numerically.

3(5x + 4y = 17)

4(4x + 3y = 12)

15x + 12y = 51 Let this be equation (iii)

12x + 12y = 48. Let this be equation (iv)

NB

When solving simultaneous equations by eliminations method.

Decide which variable to eliminate.

Make the coefficient of the variable the same in both equations.

Eliminate the variable by addition or subtraction as is appropriate.

Solve for the remaining variable

Substitute your value for part (4) above in any of the original equations to solve for the other variable.

Solving simultaneous equations by substitution method.

Writing one variable in terms of the other variables.

Find y in terms of x.

4x - y = 10

Solution

4x - y = 10

Subtract 4x form both sides of the equation.

4x - 4x -y = 10 - 4x

-y = 10 - 4x

Multiply both sides of the equation by -1

-y x - 1 + (10 - 4x) -1

y = -10 + 4x

In y = 4x - 10, y is expressed in terms of x.

b) 3x + 7y = 29

Solution

3x + 7y = 29

Subtract 3x form both sides of the equation.

3x - 3x + 7y = 29 - 3x

7y = 29 - 3x

Dividig both sides of the equation by 7.

7y = 29 - 3x

7 7

y = 29 - 3x

7

Find x in terms of y.

Subtracting 2y form both sides of the equation.

2y - 2y + x = 5 - 2y

x = 5 - 2y

c) 3x- 6y = 10

adding 6y on both sides of the equation

3x - 6y +6y = 10 + 6y

3x = 10 + 6y

Dividing both sides of the equation by 3.

3x = 10 + 6y

3 3

x = 10 + 6y

3

The processof writing a variable in terms of the other is similar

to the process of solving linear equation with one unkown.

Solve the simultaneous equation by substitution method.

4x - 3y = 1

x - 4 = 2y

Let 4x - 3y= 1 be equation (i) and

x - 4 = 2y be equation (ii)

4x - 3y = 1 -------(i)

x - 4 = 2y ------(ii)

Using equation (ii) write x in terms of y.

x - 4 + 4 = 2y + 4

x = 2y + 4 Let this be equation (iii)

Substitute 2y + 4 for x in equation i)

4(2y + 4) - 3y = 1

8y + 16 - 3y = 1

8y - 3y + 16 = 1

5y + 16 = 1

5y + 16 - 16 = 1 16

5y = -15

5y 5

y = -3

Substituting -3 for y in either equations (i) or (ii) to get the value of x.

Using equation (i)

4x - 3y = 1

4x - 3(-3) = 1

4x -(-9) = 1

4x + 9 =1

4x + 9 - 9 = 1 -9

4x = -8

4 4

x = -2

Using Equation (ii)

x - 4 = 2y

x - 4 = 2(-3)

x - 4 = -6

x - 4 + 4 = -6 + 4

x = -2

If an equation is used to write one variabe in terms of the other, it cannot be used again for substitution.

Solve the simultaneous linear equations by substitution.

2t + 3s = 8

6t + 5 s = 16

Let 2t + 3s = 8 be equation (i) and

6t + 5s = 16 be equation (ii)

Using equation (i) write in terms of s

2t + 3s - 3s = 8 - 3s

2t = 8-3s

2 2

t = 8-3s

2

Let this equation be equation (iii)

Using equation (i) write t in terms of s.

Substitute 8-3s for t in equation (ii)

2

6(8-3s) + 5s = 16)2

2

Multiply both sides of the equation by 2.

6(8 - 3s) + 10S = 32

48 - 18S + 10S = 32

48 - 48 - 8S = 32 48

-8S = -16

-8 -8

S = 2.

Substituting 2 for s in equation either equation i, ii or iii to get the value of t in equation (iii)

t = 8 - 3(2)

2

t = 8 - 6

2

t = 2

2

t = 1

Solve the simultaneous linear equations by substitution method.

5x + 2y = 10

3y + 7x = 29

Let 5x + 2y be equation (i) and

3y + 7x = 29

5x + 2y = 10 -----------------(i)

3y + 7x = 29-----------------(i

Using Equation (i) write x in terms of y.

5x + 2y = 10

5x + 2y - 2y + 10 - 2y

5x = 10 - 2y

Divide both sides by 5

5x = 10 - 2y

5 5

x = 10 - 2y

5

Substituting x = 10 - 2y

5

for x in equation (ii)

3y + 7(10 - 2y)

= 29 5

Multiply both sides by 5.

(3y + 7(10 - 2y)

29) 5 5

15y + 7(10 - 2y) = 145

15y + 70 - 14y = 145

15y - 14y + 70 = 145

y + 70 = 145

y + 70 - 70 = 145 70

y = 75

Substitute 75 for y in equation (iii)

x = 10 - 2(75)

5

x = 10 - 150

5

x = -140

5

x = -28

Elimination method.

Solving simultaneous equations by elimination method.

3x - 2y = 11

2x - 2y = 10

Method

Let 3x - 2y = 11 be equation (i

and 2x - 2y = 10 be equation (ii)

Substract equation (ii) from equation (i)

3x - 2y = 11

- 2x - 2y =10

x = 1

The Value of x is now equal to 1. In equation (i) or (ii) substitute 1 for x.

Take3x - 2y = 11

3x1 - 2y = 11

3 - 2y = 11

3 - 3 - 2y = 11 - 3

-2y = 9

-2y = 8 -2 -2

y = -4

The values of x and y that satisfies both simultaneous linear equations are

x = 1 and y = -4.

Solve the simultaneous equation

2x - 4y = 2

12x + 4y = 40

Let 2x - 4y = 2 be equation (i)

and 12x + 4y = 40 be equation (ii)

Add equation (i) to equation (ii)

2x - 4y = 2

+ 12x + 4y = 40

14x = 42

Solve this linear equation by dividing both sides of the equation by 14.

14x = 42 14 14

x =3

The value of x is 3.

In equation (i) and (ii) substitute 3 for x.

Take 2x - 4y = 2

2x3 - 4y = 2

4 - 4y = 2

5 - 6 - 4y = 2 -6

-4y = -4 -4 -4

y = 1

The values of x and y that satisfies both simultaneous linear equations

are x = 3 and y = 1

In the above two illustrations, it was easy to eliminate one unknown

for its coefficients were numerically equal. When the signs infront of the

unknown to be eliminated are the same we subtract, and when the signs

are different we add.

Solve the simultaneous equations.

9x + 3y = 21

3x - 6y = 0

Let 9x + 3y = 21 be equation (i)

and 3x - 6y =0 be equation (ii)

Note the coefficients of the unknowns are not the same.

Multiply equation (ii) by 3 to make the coefficients of x equal numerically.

(3x - 6y = 0) x3

9x - 18y = 0

Let this equation be equation (iii)

9x - 18y = 0 ----------(iii)

Subtract equation (iii) from equation (i) to eliminate x.

9x + 3y = 21 --------(i)

9x - 18y = 0 ---------(ii)

21y = 21

Dividing both sides by 21

21y = 21 21 21

y = 1

In equations (i), (ii) or (iii) substitute 1 for y.

9x + 3y = 21

9x + 3 x 1 = 21

9x + 3 = 21

9x + 3 - 3 = 21 3

9x = 18

9x = 18 9 9

x = 2

The values of x and y that satisfies the equations are x = 2 and y = 1

Solve the simultaneous linear equations

5x + 4y = 17

3x + 3y = 12

Let 5x + 4y = 17 be equation (i)

and 3x + 3y = 12 be equation (ii)

5x + 4y = 17 --------- (i)

3x + 3y = 12 --------- (ii)

Multiply equation (i) by 3 and equation (ii) by 5, so as to make

the coefficients of x equal numerically.

3 (5x + 4y = 17)

15x + 12y = 51. Let this be equation (iii)

and 5(3x + 3y = 12)

15x + 15y = 60. Let this be equation (iv)

Subtract equation (iv) form equatiom (iii) to eliminate x now that its

coefficients are numerically equal in both equations.

15x + 12y = 51----------(iii)

- 15x + 15y = 60 ---------- (iv)

-3y = -9

-3y = -9 -3 -3

y = 3

In either equations (i), (ii), (iii) or (iv) substitute 3 for y to solve for x.

Taking 3x + 3y = 12

3x + 3(3) = 12

3x + 9 = 13

3x + 9 - 9 = 12 9

3x = 3

x = 1

Note in solving the simultaneous equation above, that is

15x + 12y = 51----------(iii)

- 15x + 15y = 60 ---------- (iv)

Multiply equation (i) by 3 and equation (ii) by 4 to make the coefficients

of y equal numerically.

3(5x + 4y = 17)

4(4x + 3y = 12)

15x + 12y = 51 Let this be equation (iii)

12x + 12y = 48. Let this be equation (iv)

When solving simultaneous equations by eliminations method.

Decide which variable to eliminate.

Make the coefficient of the variable the same in both equations.

Eliminate the variable by addition or subtraction as is appropriate.

Solve for the remaining variable

Substitute your value for part (4) above in any of the original equations

to solve for the other variable.

BACKGROUND KNOWLEDGE

Natural numbers

The first number we ever meet are the whole numbers, also called natural numbers,and these are written down using numerals.

Numerals and Place value

The whole numbers ornatural numbers are written using the ten numerals 0,1,2,...,9 where the position of a numeral dictates the value that it represents.

For example :

246 stands for 2 hundrends,4 tens and 6 units or ones.

That is 200 + 40 + 6

here the numerals 2,4 and 6 are called the hundreds,tens and units which are the place values

Composite number

A composite number is a positive integer which is not prime and not equal to 1. That is, n is composite if n = axb with a and b being natural numbers both not equal to 1.

Example:

1 is not composite (and also not prime), by definition.

2 is not composite, as it is prime.

15 is composite, since 15=3x5

93555 is composite, since 93555=3 x 3 x 3 x 3 x 3 x 5 x 7 x 11

52223 is not composite, since it is prime

Express composite numbers in factor form

This can be done in more than one way 24= 6 x 4, 8 x 3, 12 x 2, 24 x 1 24 = 2 x 2 x 2 x 3

Lesson Objective

By the end of this lesson, you should be able to;

express numbers as products of prime factors in power form correctly.

FACTORS

Factors are really quiet easy to understand.I'll explain why.

2. 54

Prime Numbers

If a natural number has only two factors which are itself and the number 1 ,the number is called a prime number.

A prime number is any number with no divisors other than itself and 1.

The first 6 prime numbers are 2,3,5,7,11 and 13.

The number one (1) is not a prime number because it only has one factor namely itself.

Any number can be written as a product of prime numbers in a unique way.e.g

6= 2 X 3

21=3 X 7

8= 2 X 2 X 2

Expressing a number as product of prime factors

Given the following numbers list down the prime numbers. 1,2,20,5,23,0.1,7,4,29,13,32. From the given list the prime number are 2, 5, 7, 13, 23, 29. We consider the number 36. The number 36 can be expressed as a product of prime factors using a tree diagram shown below:

Expressing as a product of prime factors

Given the following numbers list down the prime numbers.

1, 2, 20, 5, 23, 0.1, 7, 4, 29, 13, 32

From the given list above, the prime number are 2, 5, 7, 13, 23, 29

We consider the number 36

The number 36 can be expressed as a product of prime factors

using a tree diagram shown below:

The number 2 and 3 are the prime factors of 36 hence 36 = 2 x 2 x 3 x 3

2 x 2 can be written as 2^{2} in short form. Similarly, 3 x 3 can be written as 3^{2}

Therefore 36 = 2^{2} x 3^{2}

The number 2 in 2^{2} is called a power. It shows the number of times the number

has been multiplied by itself.

Similarly the number 2 in 3^{2} is also called a power.

2^{2} is read as "two to power two" and 3^{2} is read as "three to power two".

When we write 36 = 2^{2} x 3^{2}, we say that 36 has been expressed as a product

of prime factor in power form.

Worked out Examples

Express each of the following numbers as a product of its prime factors.

Help pageExpressing as a product of prime factors

Given the following numbers list down the prime numbers.

1, 2, 20, 5, 23, 0.1, 7, 4, 29, 13, 32

From the given list above, the prime number are 2, 5, 7, 13, 23, 29

We consider the number 36

The number 36 can be expressed as a product of prime factors

using a tree diagram shown below:

The number 2 and 3 are the prime factors of 36 hence 36 = 2 x 2 x 3 x 3

2 x 2 can be written as 2^{2} in short form. Similarly, 3 x 3 can be written as 3^{2}

Therefore 36 = 2^{2} x 3^{2}

The number 2 in 2^{2} is called a power. It shows the number of times the number has been multiplied by itself.

Similarly the number 2 in 3^{2} is also called a power.

2^{2} is read as "two to power two" and 3^{2} is read as "three to power two".

When we write 36 = 2^{2} x 3^{2}, we say that 36 has been expressed as a product of prime factor in power form.

Worked out Examples

Express each of the following numbers as a product of its prime factors.

1. 32

2. 72

3. 54

4. 225

Worked out Examples

Express each of the following numbers as a product of its prime factors.

1**.** 32

2. 54

Express each of the following numbers as a product of its prime factors.

3. 225

**Introduction**

Factors

Any pair of natural numbers are called factors of their product. For example the numbers 3 and 6 are factors of 18 because 3 x 6 = 18 .

These are not the only factors of 18.

The complete collection of factors of 18 is 1,2, 3,6, 9, and 18 because

18 = 1 x 18

18 = 2 x 9

18 = 3 x 6

If a number can be expressed as a product of two whole numbers, then the whole numbers are called factors of that number.

List down the factors of 42.

Solution:

The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

Now List down the factors of 12 Try making 12 in different ways.

Your answer should look like this: 6 x 2 = 12, 12 x 1 = 12, 4 x 3 = 12

Remember that you can write your numbers in any order you like for a multiplication:

2 x 6 is the same as 6 x 2

1 x 12 is the same as 12 x 1

3 x 4 is the same as 4 x 3.

The full list of factors of 12 is 1, 2, 3, 4, 6, and 12.

Some numbers have many factors, so it is a good idea to work in an organised way or you may miss some. Don't forget to include 1 and the number itself in your list.

Here is one way to find the factors of 48. Start with 1 and pair off your numbers.

1 x 48, 2 x 24, 3 x 16, 4 x 12 and 6 x 8 all make 48.

Write the list in order: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

Here is another way: Write your first pair of factors with a reasonable space between them, then move on to the next pair until you have them all.

1,2 3,4 6,8 12,16 24,48

Factors that are common to two or more numbers are called common factors.

Prime Numbers

If a natural number has only two factors which are itself and the number 1 ,the number is called a prime number. A prime number is any number with no divisors other than itself and 1.The first 6 prime numbers are 2,3,5,7,11 and 13.

The number one (1) is not a prime number because it only has one factor namely itself.

Any number can be written as a product of prime numbers in a unique way.e.g

6= 2 X 3

21=3 X 7

8= 2 X 2 X 2

Prime factors

Every natural number can be written as a product involving only prime factors. For example,the number 126 has the factors 1,2,3,6,7,9,14,18,21,42,63 and 126,of which 2,3 , 7 are prime numbers and 126 can be written as:

126= 2 x 3 x 3 x 7

To obtain these prime factors the number is divided by successively increasing prime number as follows:

so 126 = 2 x 3 x 3 x 7

Note that a prime factor may occur more than once. The process of writing a number in prime factors is called prime factorisation

Multiples

A multiple is a number that may be divided by another number with no remainder. For example 4,10 and 32 are multiples of 2.

Multiples of a number can be made by multiplying the number by any whole number. The first four multiples of 2 are 2, 4, 6 and 8. You get them by multiplying 2 x 1, 2 x 2, 2 x 3 and 2 x 4

The numbers you find in the 2-times table are all multiples of 2.

**Note**: when you carry out multiplication you can write the numbers in any order and get the same answer. 6 x 2 is the same as 2 x 6.

Here is how to make multiples of 10. Just multiply 10 by a whole number each time.

1 x 10 = 10,

2 x 10 = 20,

3 x 10 = 30,

4 x 10 = 40,

5 x 10 = 50,

6 x 10 = 60,

and so on ...

The first six multiples of 10 are 10, 20, 30, 40, 50 and 60.

Example 1

Is 12 a multiple of 3?

If you multiply 3 by 4 you get 12, so 12 is a multiple of 3.

Example 2

20 is a multiple of 5 because 4 x 5 = 20

20 is a multiple of 4 too, because 5 x 4 = 20.

Example 3

Is 15 a multiple of 3?

3 x 5 = 15. So 15 is a multiple of 3, (and also of 5).

Example 4

Is 21 a multiple of 6?

21 is not a multiple of 6 because you can't make 21 by multiplying 6 by any whole number.

6 x 3 = 18 and 6 x 4 = 24 but there is no whole number between 3 and 4 that could give us an answer of 21.

Example 5

Is 30 a multiple of 15? 30 = 2 x 15, so 30 is a multiple of 15. You can also see that 2 x 3 x 5 = 30 so 30 is a multiple of 2, 3 and 5. And 30 = 3 x 10 so 30 is a multiple of 10. Also 30 = 5 x 6 so 30 is a multiple of 6 too.

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