Question

Find the HCF of the following pairs of integers and express it as a linear combination of them.
(iii) 506 and 1155
(iv) 1288 and 575

Answer 1

SOLUTION :

(iii) First , we need to find the H.C.F. of 506 and 1155

Here , 1155 > 506

Let a = 1155 and b = 506

1155 = 506 x 2 + 143…………(1)

[By applying division lemma, a = bq + r]

Here, remainder = 143 ≠ 0, so take new dividend as 506 and new divisor as 43.

Let a = 506 and b= 43

506 = 143 x 3 + 77……………..(2)

[By applying division lemma, a = bq + r]

Here, remainder = 77 ≠ 0, so take new dividend as 143 and new divisor as 77.

Let a = 143 and b= 77

143 = 77 x 1 66....................(3)

[By applying division lemma, a = bq + r]

Here, remainder = 66 ≠ 0, so take new dividend as 77 and new divisor as 66.

Let a = 77 and b= 66

77 = 66 x 1 + 11………….(4)

[By applying division lemma, a = bq + r]

Here, remainder = 11 ≠ 0, so take new dividend as 66 and new divisor as 11

Let a = 66 and b= 11

66 = 11 x 6 + 0………….(5)

By applying division lemma, a = bq + r]

Here, remainder is zero and divisor is 11

Hence ,H.C.F. of 506 and 1155 is 11.

H.C.F. = 11.

Now,11 = 77 – 66 x 1

= 77- [143 -  77 x 1] x 1

[Substituting 143 -  77 x 1= 66 from eq 3]

= 77 – 143 x 1 + 77 x 1

= 77 + 77 x 1 – 143 x 1  

= 77 x 2 - 143 x 1

= [506 – 143 x 3] x 2 – 143 x 1

[Substituting 506 – 143 x 3 = 77 from eq 2]

= 506 x 2 – 143 x 6 – 143 x 1

= 506 x 2 – 143 x 7

= 506 x 2 – [1155 – 506 x 2] x 7  

[Substituting 1155 – 506 x 2 = 143 from eq 1]

= 506 x 2 – 1155 x 7+ 506 x 14

= 506 x 2 + 506 x 14 – 1155 x 7

11 = 506 x 16 – 1155 x 7

Hence the linear combination is11 = 506 x 16 - 1155 x 7

(iv) First ,we need to find the H.C.F. of 1288 and 575

Here 1288 > 575  

Let a = 1288 and b= 575

1288 = 575 x 2+ 138………..(1)

[By applying division lemma, a = bq + r]

Here, remainder = 138 ≠ 0, so take new dividend as 575 and new divisor as138.

Let a = 575 and b= 38

575 = 138 x 4 + 23…………(2)

[By applying division lemma, a = bq + r]

Here, remainder = 23 ≠ 0, so take new dividend as 138 and new divisor as 23.

Let a = 138 and b= 23

138 = 23 x 6 + 0…………..(3)

Here, remainder is zero and divisor is 23

Hence ,H.C.F. of 1288 and 575 is 23.

H.C.F. = 23.

Now, 23 = 575 – 138 x 4  

= 575 - [1288 – 575 x 2] x 4

[Substituting 1288 – 575 x 2 = 138 from eq 1]

= 575 – 1288 x 4 + 575 x 8

= 575 + 575 x 8 – 1288 x 4

23 = 575 ×9 + (-4) × 1288

Hence, the linear combination is  23 = 575 ×9 + (-4) × 1288.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Hope it's helpful...

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