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
