Find HCF of 81 and 237 and express it as a linear combination of 81 and 237

Since, 237 > 81

On applying Euclid’s division algorithm, we get

237 = 81 × 2 + 75 …(i)

81 = 75 × 1 + 6 …(ii)

75 = 6 × 12 + 3 …(iii)

6 = 3 × 2 + 0 …(iv)
Hence, and HCF (81, 237) = 3. 1 Write 3 in the form of 81x + 237y, move backwards
3 = 75 – 6 × 12 [From (iii)]
= 75 – (81 – 75 × 1) × 12 [Replace 6 from (ii)]
= 75 – (81 × 12 – 75 × 1 × 12)
= 75 – 81 × 12 + 75 × 12
= 75 + 75 × 12 – 81 × 12
= 75 ( 1 + 12) – 81 × 12
= 75 × 13 – 81 × 12
= 13(237 – 81 × 2) – 81 × 12 [Replace 75 from (i)]
= 13 × 237 – 81 × 2 × 13 – 81 × 12
= 237 × 13 – 81 (26 + 12)
= 237 × 13 – 81 × 38
= 81 × (– 38) + 237 × (13)
= 81x + 237y
Hence, x = – 38 and y = 13

Hope that helps :slight_smile:

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