What is common factors?

Example• Factor of 6 are 1 2 3 6• Factor of 18 are 1 2 3 6 9 and

18

You will see that the factors 1, 2, 3 and 6 are COMMON FACTORS of 6 and 18.

And from the common factors you will see that the HIGHEST COMMON FACTORS (H.C.F.) of 6 and 18 is 6

Another method

We can also find the HCF by using Prime Factorization method which is the easier and faster way.

Using Prime Factorization

Find the highest common factor of 20 and 30 20 = 2 x 2 x 5

30 = 2 x 3 x 5 2 x 5 = 10

Therefore, the highest common factors of 20 and 30 is 10

Using Prime Factorization

Find the highest common factor of 20 and 30 20 = 2 x 2 x 5

30 = 2 x 3 x 5 2 x 5 = 10

Therefore, the highest common factors of 20 and 30 is 10

Using Repeated Division

Find the highest common factor of 20 and 30 2 20 30

5 10 15 2 3

HCF is 2 x 5 = 10Therefore, the highest common factors of 20 and 30 is 10

Practice 1

1. Find the highest common factor of 24, 60, 84 (Using Prime Factorization)

2. Find the highest common factor of 32, 48, 104 (Using Repeated Division)

3. Find the highest common factor of 24, 36, 60 4 .Find the highest common factor of 42, 63, 105

Uncle samarn has 3 robes with the lengths of 84, 108 and 156 meters. If he wants to each robe into shorter ones of the same lengths at the maximum possible length for each, how long will each robe be and how many robes will he get?

Example

Solution The length of each small robe can be determined by the biggest number that can divide 84, 108 and 156 exactly.

Therefore, we need to find the G.C.D. of 84, 108 and 156

84 108 15642 54 7821 27 39 7 9 13

The G.C.D of 84, 108 and 156 is 2 x 2 x 3 = 12The length of each robe is 12 meters.

2

2

3

Uncle samarn has 3 robes with the lengths of 84, 108 and 156 meters. If he wants to each robe into shorter ones of the same lengths at the maximum possible length for each, how long will each robe be and how many robes will he get?

Example

The first robe is 84 meters. Cut into shorter ones with the length of 12 meters will get 84/12 = 7 robes.

The second robe is 108 meters. Cut into shorter ones with the length of 12 meters will get 108/12 = 9 robes.

The second robe is 156 meters. Cut into shorter ones with the length of 12 meters will get 156/12 = 13 robes.

He will get 7+ 9 + 13 = 29 robes in total.Therefore, each robe will be 12 meters long and

he will get 29 robes.

Sommai has 38 pigs, 86 chickens and 122 cows, He wants to divide the pigs, chickens and cows into groups; each group has the same amount of animals and the largest amount possible. Morever, each group will have only 1 type of animal and after the division there should be only 2 animals of each type left. How many group of animals can he get and how many animals in each group?

Example

Find the maximum number that after dividing 218 and 371 will leave 2 as the remainder

Example