#### Question

Find the HCF of the following pairs of integers and express it as a linear combination of 506 and 1155.

#### Solution

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 + 56 ….(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 36 and remainder 9

66 = 11 × 6 + 0

∴ HCF = 11

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 – 115 × 7