Find the HCF of the following pair of integers and express it as a linear combination of them i 963 ...

Question

Find the HCF of the following pair 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

Accepted Answer

By applying Euclids 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 Thus we can conclude the HCF 9 Now in order 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 Euclids 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 Thus we can conclude the HCF 4 Now in order 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 Euclids 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 Thus we can conclude the HCF 11 Now in order 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 Euclids 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 Thus we can conclude the HCF 23 Now in order 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

RD Sharma Solutions Class 10 Mathematics Solutions for Real Numbers Exercise 1.2 in Chapter 1 - Real Numbers