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Question

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

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Solution

We need to find the H.C.F. of 963 and 657 and express it as a linear combination of 963 and 657. By applying Euclid’s division lemma, 963 = 657 x 1 + 306.

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

657 = 306 x 2 + 45.

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

306 = 45 x 6 + 36.

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

45 = 36 x 1 + 9.

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

36 = 9 x 4 + 0.

Therefore, H.C.F. = 9.

Now, 9 = 45 – 36 x 1

= 45 – [306 – 45 x 6] x 1 = 45 – 306 x 1 + 45 x 6

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

= 657 x 7 – 306 x 14 – 306 x 1

= 657 x 7 – 306 x 15

= 657 x 7 – [963 – 657 x 1] x 15

= 657 x 7 – 963 x 15 + 657 x 15

= 657 x 22 - 963 x 15

Hence, obtained.

(ii) We need to find the H.C.F. of 592 and 252 and express it as a linear combination of 592 and 252.

By applying Euclid’s division lemma

592 = 252 x 2 + 88

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

252 = 88 x 2 + 76

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

88 = 76 x 1 + 12

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

76 = 12 x 6 + 4

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

12 = 4 x 3 + 0.

Therefore, H.C.F. = 4

Now, 4 = 76 – 12 x 6

= 76 – 88 – 76 x 1 x 6

= 76 – 88 x 6 + 76 x 6

= 76 x 7 – 88 x 6

= 252 – 88 x 2 x 7 – 88 x 6

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

= 252 x 7- 88 x 20

= 252 x 7 – 592 – 252 x 2 x 20

= 252 x 7 – 592 x 20 + 252 x 40

= 252 x 47 – 592 x 20

= 252 x 77 + 592 x (-20)

Hence obtained.

(iii) We need to find the H.C.F. of 506 and 1155 and express it as a linear combination of 506 and 1155. By applying Euclid’s division lemma

1155 = 506 x 2 + 143.

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

506 = 143 x 3 + 77.

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

143 = 77 x 1 + 66.

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

77 = 66 x 1 + 11.

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

66 = 11 x 6 + 0.

Therefore, H.C.F. = 11.

Now, 11 = 77 – 66 x 1 = 77—[143 – 77 x 1] x 1

= 77 – 143 x 1 + 77 x 1

= 77 x 2 – 143 x 1

= [506 – 143 x 3] x 2 – 143 x 1

= 506 x 2 – 143 x 6 – 143 x 1

= 506 x 2 – 143 x 7 = 506 x 2 – [1155 – 506 x 2] x 7 = 506 x 2 – 1155 x 7+ 506 x 14

= 506 x 16 - 1155 x 7

Hence obtained.

(iv) We need to find the H.C.F. of 1288 and 575 and express it as a linear combination of 1288 and 575. By applying Euclid’s division lemma

1288 = 575 x 2+ 138.

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

575 = 138 x 4 + 23.

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

138 = 23 x 6 + 0.

Therefore, H.C.F. = 23.

Now, 23 = 575 – 138 x 4 = 575 - [1288 – 575 x 2] x 4

= 575 - 1288 x 4 + 575 x 8

= 575 x 9 - 1288 x 4

Hence, obtained.


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