Math, asked by ooPriyanshu7195, 9 months ago

Find the HCF of the following pairs of integers and express it as a linear combination of them
(i) 963 & 657 (ii) 592 & 252
(iii) 506 & 1155 (iv) 1288 & 575

Answers

Answered by topwriters
39

HCF is (i) 9 (ii) 4 (iii) 11 (iv) 23

Step-by-step explanation:

(i) 963 & 657

By applying Euclid’s division lemma on 963 and 657, we get

963 = 657 x 1 + 306 ---- (1)

As the remainder ≠ 0, apply division lemma on divisor 657 and remainder 306

657 = 306 x 2 + 45 ---- (2)

As the remainder ≠ 0, apply division lemma on divisor 306 and remainder 45

306 = 45 x 6 + 36 ---- (3)

As the remainder ≠ 0, apply division lemma on divisor 45 and remainder 36

45 = 36 x 1 + 9 ---- (4)

As the remainder ≠ 0, apply division lemma on divisor 36 and remainder 9

36 = 9 x 4 + 0 ---- (5)

Therefore H.C.F. = 9.

To express the found HCF as a linear combination of 963 and 657, we perform

9 = 45 – 36 x 1 [from (5)]

= 45 – [306 – 45 x 6] x 1 = 45 – 306 x 1 + 45 x 6 [from (3)]

= 45 x 7 – 306 x 1 = [657 -306 x 2] x 7 – 306 x 1 [from (2)]

= 657 x 7 – 306 x 14 – 306 x 1

= 657 x 7 – 306 x 15

= 657 x 7 – [963 – 657 x 1] x 15 [from (1)]

= 657 x 7 – 963 x 15 + 657 x 15

= 657 x 22 – 963 x 15.

(ii) 592 and 252

By applying Euclid’s division lemma on 592 and 252, we get

592 = 252 x 2 + 88 ---- (1)

As the remainder ≠ 0, apply division lemma on divisor 252 and remainder 88

252 = 88 x 2 + 76 ---- (2)

As the remainder ≠ 0, apply division lemma on divisor 88 and remainder 76

88 = 76 x 1 + 12----(3)

As the remainder ≠ 0, apply division lemma on divisor 76 and remainder 12

76 = 12 x 6 + 4 ----(4)

Since the remainder ≠ 0, apply division lemma on divisor 12 and remainder 4

12 = 4 x 3 + 0 ----(5)

Therefore H.C.F. = 4.

To express the found HCF as a linear combination of 592 and 252, we perform

4 = 76 – 12 x 6 [from (4)]

= 76 – [88 – 76 x 1] x 6 [from (3)]

= 76 – 88 x 6 + 76 x 6

= 76 x 7 – 88 x 6

= [252 – 88 x 2] x 7 – 88 x 6 [from (2)]

= 252 x 7- 88 x 14- 88 x 6

= 252 x 7- 88 x 20

= 252 x 7 – [592 – 252 x 2] x 20 [from (1)]

= 252 x 7 – 592 x 20 + 252 x 40

= 252 x 47 – 592 x 20

= 252 x 47 + 592 x (-20)

(iii) 506 and 1155

By applying Euclid’s division lemma on 506 and 1155, we get

1155 = 506 x 2 + 143 ---- (1)

As the remainder ≠ 0, apply division lemma on divisor 506 and remainder 143

506 = 143 x 3 + 77 ---- (2)

As the remainder ≠ 0, apply division lemma on divisor 143 and remainder 77

143 = 77 x 1 + 66 ---- (3)

Since the remainder ≠ 0, apply division lemma on divisor 77 and remainder 66

77 = 66 x 1 + 11 ---- (4)

As the remainder ≠ 0, apply division lemma on divisor 66 and remainder 11

66 = 11 x 6 + 0 ---- (5)

Therefore H.C.F. = 11.

To express the found HCF as a linear combination of 506 and 1155, we perform

11 = 77 – 66 x 1 [from (4)]

= 77 – [143 – 77 x 1] x 1 [from (3)]

= 77 – 143 x 1 + 77 x 1

= 77 x 2 – 143 x 1

= [506 – 143 x 3] x 2 – 143 x 1 [from (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 [from (1)]

= 506 x 2 – 1155 x 7+ 506 x 14

= 506 x 16 – 1155 x 7

(iv) 1288 and 575

By applying Euclid’s division lemma on 1288 and 575, we get

1288 = 575 x 2+ 138 ---- (1)

As the remainder ≠ 0, apply division lemma on divisor 506 and remainder 143

575 = 138 x 4 + 23 ---- (2)

As the remainder ≠ 0, apply division lemma on divisor 143 and remainder 77

138 = 23 x 6 + 0 ---- (3)

Therefore H.C.F. = 23.

To express the found HCF as a linear combination of 1288 and 575, we perform

23 = 575 – 138 x 4 [from (2)]

= 575 – [1288 – 575 x 2] x 4 [from (1)]

= 575 – 1288 x 4 + 575 x 8

= 575 x 9 – 1288 x 4

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