Find the HCF of the following pairs of integers and express it as a linear combination of them. (i) 963 and 657

(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.