3. The lowest common multiple
The lowest common multiple (or LCM) of two numbers is the smallest number that is a multiple of both the numbers.
For small numbers we can find this by writing down the first few multiples for both numbers until we find a number that is in both lists.
For example,
Multiples of 20 are :
20,
40,
60,
80,
100,
120, . . .
Teacher notes
You may like to add that if the two numbers have no common factors (except 1) then the lowest common multiple of the two numbers will be the product of the two numbers.
For example, 4 and 5 have no common factors, so the lowest common multiple of 4 and 5 is 4 × 5 = 20.
Pupils could also investigate this themselves later in the lesson.
Multiples of 25 are :
25,
50,
75,
100,
125, . . .
The LCM of 20 and 25 is: 100.

4. LCM
Teacher notes
This activity will test the students’ ability to correctly recognize multiples of numbers. There are 2 draggable items that contain the number 24 in this activity. If they are both placed in one column, the activity can not be completed ‘correctly’. Although technically correct, the other answers will not be able to all be right, so the activity as a whole will be incorrect. The idea is that the students should realize that 24 is a multiple of 8 and 12. More than that, they should also be able to work out that it is the LCM of 8 and 12.

5. More LCM

6. The lowest common multiple
Teacher notes
It is possible to stop the animation at any point by pressing the pause button. Logical chapter points have been added at points where you may wish to prompt the students to think about the next stage. You may wish to remind pupils that to add two fractions together they must have the same denominator. The LCM is the lowest number that both 8 and 12 will divide into. Ask pupils how many 8s ‘go into’ 24. Establish that we must multiply 8 by 3 to get 24.
Remind the students that once we’ve multiplied the bottom by 4, so we must multiply the top by 4. If you multiply the top and the bottom of a fraction by the same number, you do not change its value. 15/24 is just another way of writing 5/8. Repeat this explanation as you convert 7/12 to 14/24.
15/24 plus 14/24 equals 29/24. Can this fraction be simplified? Establish that it can be converted to a mixed number. We can represent 29/24 as 1 5/24

8. The highest common factor
The highest common factor (or HCF) of two numbers is the highest number that is a factor of both numbers.
We can find the highest common factor of two numbers by writing down all their factors and finding the largest factor in both lists.
For example,
Factors of 36 are : 1, 2, 3, 4, 6, 9,
12,
18,
36.
Teacher notes
Point out that 3 is a common factor of 36 and 45. So is 1. But 9 is the highest common factor.
Factors of 45 are : 1, 3, 5, 9,
15,
45.
The HCF of 36 and 45 is: 9.

9. HCF

10. More HCF

11. The highest common factor
We use the highest common factor when cancelling fractions. 36 48
For example, can you cancel the fraction ?
The HCF of 36 and 48 is 12.
We need to divide the numerator and the denominator by 12.
The numerator cancels to: 36 ÷ 12 = 3.
The denominator cancels to: 48 ÷ 12 = 4.
÷12 36 48 3 = 4
Teacher notes
Talk through the use of the highest common factor to cancel fractions in one step.
÷12 36 48 4 3
The fraction cancels to

13. Prime factor decomposition
Teacher notes
Factor tree method: Explain that to write 36 as a product or prime factors we start by writing 36 at the top (of the tree). Next, we need to think of two numbers which multiply together to give 36: in this example we are using 9 and 4. Explain that it doesn’t matter whether we use: 2 × 18, 3 × 12, 4 × 9, or 6 × 6, the end result will be the same.
Next, we must find two numbers that multiply together to make 4. 2 is a prime number so we can stop there. Now find 2 numbers which multiply together to make 9. 3 is a prime number so we can stop there.
State that when every number at the bottom of each branch is circled we can write down the prime factor decomposition of the number writing the prime numbers in order from smallest to biggest.
Explain that it’s usual to use index notation for this, particularly with large numbers.
Repeated division by prime factors: Start by writing 36 in the table. What is the lowest prime number that divides into 36? Write the 2 to the left of the 36 and then divide 36 by 2. We write the result of this division under the 36. What is the lowest prime number that divides into 18? Continue dividing by the lowest prime number possible until you get to 1. When you get to 1 at the bottom, stop.
The prime factor decomposition is found by multiplying together all the numbers in the left hand column. Again, this is usually represented using index notation.

14. Prime factor decomposition
We can use prime factor decomposition to find the HCF and the LCM of larger numbers.
Any number can be written as a product of prime numbers.
Can you write 60 as a product of prime numbers? 2 60 2 30
Start with the smallest prime number that divides into 60.
Now use the smallest prime number that divides into 30.
Follow the same technique for 15.
The final stage sees us divide 5 by 5. 3 15 5 5
Teacher notes
This is only one method of writing numbers as a product of prime numbers. Later slides also look at prime factor trees.
Once you have established that 60, when written as a product of prime numbers, is equivalent to 2 × 2 × 3 × 5, you may wish to introduce the idea of using indices to further simplify: 60 = 2² × 3 × 5. 1
60 = 2 × 2 × 3 × 5

15. Prime factor decomposition
Teacher notes
As an extension activity, you may wish to ask students to write the numbers as a product of prime numbers using index notation whenever possible.

16. Prime factor trees
Another method of writing a number as a product of prime factors is by using a prime factor tree.
Can you write 294 as a product of prime factors?
294 2
147
Teacher notes
At this stage, you may wish to ask students if they could branch this tree differently? Students don’t have to always use the lowest multiple when constructing a factor tree, the important thing is that they divide each branch as fully as possible. If they do this, regardless of the route they take, they will end up with the same answer. Demonstrate this with the example on the slide. Instead of using 2 and 147 as the first branches, ask students to identify two other numbers that are factors of This could be 6 and 49 or 14 and 21. 3 49
294 = 2 × 3 × 7 × 7 7 7

17. Prime factor trees

18. Finding the HCF and LCM
Writing larger numbers as the product of prime factors can
make it easier to calculate the HCF and LCM of two numbers.
What is the HCF and the LCM of 60 and 294?
60 = 2 × 2 × 3 × 5
294 = 2 × 3 × 7 × 7 60
294 7 2 2 5 3 7
Teacher notes
We can find the HCF and LCM by using a Venn diagram. We put the prime factors of 60 in the first circle. Any factors that are common to both 60 and 294 go into the overlapping section. Click to demonstrate this. Point out that we can cross out the prime factors that we have included from 294 in the overlapping section to avoid adding then twice. We put the prime factors of 294 in the second circle. The prime factors which are common to both 60 and 294 will be in the section where the two circles overlap.
To find the highest common factor of 60 and 294 we need to multiply together the numbers in the overlapping section.
The lowest common multiple is found by multiplying together all the prime numbers in the diagram.
The HCF of 60 and 294 =
2 × 3
= 6.
The LCM of 60 and 294 =
2 × 5 × 2 × 3 × 7 × 7 =
2940.

19. Finding the HCF and LCM
ACTIVITY NOT CURRENTLY WORKING – NEEDS FIXED SO THAT IT WORKS ON RESET.
Teacher notes
Ask volunteers to take turns to find the LCM and HCF using the Venn diagram.

20. Using factor trees to find LCM
It is also possible to calculate the LCM of two numbers using prime factor trees.
What is the LCM of 54 and 40? 54 40 2 2 27 20 3 9 2 10 3 3 2 5
54 = 2 × 3 × 3 × 3
40 = 2 × 2 × 2 × 5
The LCM of 52 and 40 =
2 × 3 × 3 × 3 × 2 × 2 × 5 =
1080. 20

21. Using factor trees to find HCF
Prime factor trees can also be used to calculate the HCF of two numbers.
What is the HCF of 90 and 375? 90
375 2 3 45
125 3 5 15 25 3 5 5 5
90 = 2 × 3 × 3 × 5
375 = 3 × 5 × 5 × 5
The HCF of 90 and 375 =
3 × 5 =
15. 21

22. Problem solving
When numbers are expressed as a product of their prime factors, we can work out their values even if they are hidden from us in the initial expression.
What are the values of a, b and c in this expression?
a²b²c = 396, where a, b and c are all prime numbers.
Teacher notes
Students need to complete a prime factor decomposition on the number 396.
396 = 2 × 2 × 3 × 3 × 11. Therefore a = 2, b = 3 and c = 11.
Peter is incorrect. Just because a number is bigger, doesn’t mean it has a greater number of prime factors. As an example, students should be able to choose a large number that has only two prime factors, such as Its prime factors are: 1279 × 1907.
Contrast this with the number which has three prime factors. Its prime factors are: 26 × 56 × 17
Peter thinks that the larger the number, the greater the number of prime factors. Is he correct? Show your working and include examples to support your claim.

24. A snail’s pace
Two snails are being raced around a circular track. It takes one snail 4 minutes to complete a full lap and the other snail takes 7 minutes to complete one lap.
How long will it be before they next meet at the starting point?
A third snail takes 3 minutes to complete a lap.
Teacher notes
It takes the first snail 4 minutes to complete one lap. Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, …
It takes the second snail 7 minutes to complete one lap. Multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, …
The LCM of 4 and 7 is 28. The first two snails will meet after 28 minutes.
The third snail takes 3 minutes to complete a lap. The LCM of 3, 4 and 7 is: 2² × 3 × 7 = 84. This means that 1 hour and 24 minutes will have passed before all three snails will meet up again at the starting point.
By this time:
snail one will have completed 84 ÷ 4 = 21 laps
snail two will have completed 84 ÷ 7 = 12 laps
snail three will have completed 84 ÷ 3 = 28 laps.
How long will it be before all three snails are at the starting point again?
How many laps will each snail have completed?

25. Rolling coins
Two coins with circumferences of 6 cm and 8 cm are rolled along a flat surface.
What is the shortest distance the coins can roll if they both have to travel an exact number of turns?
Why can the coins not travel a distance of 56 cm together in an exact number of turns?
Investigate what the shortest distance is that two different coins can travel before completing an exact number of turns.
Teacher notes
Students should solve this problem using the LCM of the two coins.
Multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, …
Multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64, …
Alternatively, 6 = 2 × 3, 8 = 2³. The LCM of 6 and 8 is: 2 × 3 × 2 × 2 = 24.
The LCM of 6 and 8 is 24. Therefore, the shortest distance the coins can roll before they have both turned an exact number of revolutions is 24 cm. After 24 cm, the 8 cm coin will have completed 3 revolutions and the 6cm coin will have completed 4 revolutions.
The coins can not travel 56 cm in an exact number of turns as 56 is not a common multiple of 6 and is only a multiple of 8.
Encourage students to use real coins for their investigations. They could use 1 p, 2 p, 5 p, 10 p and £1 coins. Alternatively, you could get them to take in different currencies and compare and contrast their sizes.

26. Flower power A garden designer is planning his rectangular flowerbeds.
The first bed will feature red and yellow flowers and the display must be as wide as possible.
Each row can only contain one colour and all plants must be used.
If he has 60 red plants and 36 yellow plants how wide will the beds be?
In another bed he has 3 different coloured flowers. Using the same approach, how wide will the beds be if he has 105 red, 60 yellow and 150 white plants?
Teacher notes
We need to calculate the HCF of 60 and 36. This can be done by doing a prime factor decomposition on both numbers.
60 = 2² × 3 × 5
36 = 2² × 3²
The HCF of 60 and 36 is: 2² × 3 = 12.
Therefore, the beds will be 12 units wide and there will be 5 rows of 12 red plants and 3 rows of 12 yellow plants.
105 = 3 × 5 × 7
150 = 2 × 3 × 5²
The HCF of 60, 105 and 150 is 3 × 5 = 15.
Therefore the beds will be 15 units wide and there will be 7 rows of red flowers, 4 rows of yellow flowers and 10 rows of white flowers.