(i) 963 and 657
By applying Euclid’s division lemma 963 = 657 × 1 + 306 …(i)
Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306
657 = 306 × 2 + 45 ….. (ii)
Since remainder ≠ 0, apply division lemma on divisor 306 and remainder 45
306 = 45 × 6 + 36 …..(iii)
Since remainder ≠ 0, apply division lemma on divisor 45 and remainder 36
45 = 36 × 1 + 9 …… (iv)
Since remainder ≠ 0, apply division lemma on divisor 36 and remainder 9
36 = 9 × 4 + 0
∴ HCF = 9
For linear combination,
Now 9 = 45 – 36 × 1 [from (iv)]
= 45 – [306 – 45 × 6] × 1 [from (iii)]
= 45 – 306 × 1 + 45 × 6
= 45 × 7 – 306 × 1
= 657 × 7 – 306 × 14 – 306 × 1 [from (ii)]
= 657 × 7 – 306 × 15
= 657 × 7 – [963 – 657 × 1] × 15 [from (i)]
= 657 × 22 – 963 × 15
= 657 x 22 - 963 x 15 which is required linear combination.
(ii) 595 and 252
By applying Euclid’s division lemma
595 = 252 × 2 + 91 ….. (i)
Since remainder ≠ 0, apply division lemma on divisor 252 and remainder 91
252 = 91 × 2 + 70 …. (ii)
Since remainder ≠ 0, apply division lemma on divisor 91 and remainder 70
91 = 70 × 1 + 21 ….(iii)
Since remainder ≠ 0, apply division lemma on divisor 70 and remainder 21
70 = 21 × 3 + 7 …..(iv)
Since remainder ≠ 0, apply division lemma on divisor 21 and remainder 7
21 = 7 × 3 + 0
H.C.F = 7
For linear combination,
Now, 7 = 70 – 21 × 3 [from (iv)]
= 70 – [90 – 70 × 1] × 3 [from (iii)]
= 70 – 91 × 3 + 70 × 3
= 70 × 4 – 91 × 3
= [252 – 91 × 2] × 4 – 91 × 3 [from (ii)]
= 252 × 4 – 91 × 8 – 91 × 3
= 252 × 4 – 91 × 11
= 252 × 4 – [595 – 252 × 2] × 11 [from (i)]
= 252 × 4 – 595 × 11 + 252 × 22
= 252 × 26 – 595 × 11
Which is required linear combination.
(iii) 506 and 1155
By applying Euclid’s division lemma
1155 = 506 × 2 + 143 …. (i)
Since remainder ≠ 0, apply division lemma on division 506 and remainder 143
506 = 143 × 3 + 77 ….(ii)
Since remainder ≠ 0, apply division lemma on division 143 and remainder 77
143 = 77 × 1 + 66 ….(iii)
Since remainder ≠ 0, apply division lemma on division 77 and remainder 66
77 = 66 × 1 + 11 …(iv)
Since remainder ≠ 0, apply division lemma on divisor 66 and remainder 11
66 = 11 × 6 + 0
∴ HCF = 11
For linear combination
Now, 11 = 77 – 6 × 11 [from (iv)]
= 77 – [143 – 77 × 1] × 1 [from (iii)]
= 77 – 143 × 1 – 77 × 1
= 77 × 2 – 143 × 1
= [506 – 143 × 3] × 2 – 143 × 1 [from (ii)]
= 506 × 2 – 143 × 6 – 143 × 1
= 506 × 2 – 143 × 7
= 506 × 2 – [1155 – 506 × 27 × 7] [from (i)]
= 506 × 2 – 1155 × 7 + 506 × 14
= 506 × 16 – 1155 × 7
Which is required linear combination.
(iv) 1288 and 575
By applying Euclid’s division lemma
1288 = 575 × 2 + 138 …(i)
Since remainder ≠ 0, apply division lemma on division 575 and remainder 138
575 = 138 × 4 + 23 …(ii)
Since remainder ≠ 0, apply division lemma on division 138 and remainder 23
138 = 23 x 6 + 0
∴ HCF = 23
For linear combination
Now, 23 = 575 – 138 × 4 [from (ii)]
= 575 – [1288 – 575 × 2] × 4 [from (i)]
= 575 – 1288 × 4 + 575 × 8
= 575 × 9 – 1288 × 4
Which is required linear combination.
