Question

Q1) find the HCF of the following. à)12 , 16 and 28. (maths)​

Answer 1

Answer:

The HCF of 12, 16 and 28 is 4

Hope it helps you:)

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Now, we have the numbers 12, 16 and 28.

We will take the numbers and 12 and 16 first and find their HCF.

Now, according to Euclid’s Division Lemma,

a = bq + r , 0 < r < b.

We apply this until we get the remainder ‘r’ as ‘0’ taking ‘b’ and ‘r’ as the next ‘a’ and ‘b’ respectively and the value of ‘b’ for which the value of ‘r’ is 0 will be the required HCF.

Here, 16>12 , therefore, a=16 and b=12

Applying the division lemma to 16 and 12 we get:

16=12(1)+4 (here, q=1)

Since, r=4≠0, we will apply the division lemma again for 12 and 4.

Applying Euclid’s division lemma for 12 and 4 we get:

12=4(3)+0(here, q=3)

Here, r=0 and b=4 .

Thus, the HCF of 12 and 16 is ‘4’.

Now that we have the HCF of ’12’ and ‘16’ as ‘4’, we can find the HCF of ‘4’ and ‘28’ and it will be our required HCF.

Here, 28>4 , therefore, a=28 and b=4

Applying Euclid’s Division Lemma for 4 and 28 we get

28=4(7)+0(here, q=7)

Here, r=0 and b=4.

Thus, the HCF of ‘4’ and ‘28’ is ‘4’.

Note: This question can also be done by the following method:

We can factorise 12, 16 and 28 separately and then multiply all their prime factors. Hence, we will get our required HCF.

Factorisation of 12 is shown below:

2 | 12

2 | 6

3 | 3

1 | 1

Thus, 12 can be written as:

12=22×3

Factorisation of 16 is shown below:

2 | 16

2 | 8

2 | 4

2 | 2

1 | 1

Thus, 16 can be written as:

16=2⁴

Factorisation of 28 is shown below:

2 | 28

2 | 14

7 | 7

1 l 1

Thus, 28 can be written as:

28=2²×7

Here, we can see that 2² is a common factor for all the three numbers.

Thus, HCF=2²=4