Question

Hello tricky Problem

Find HCF of 81 and 237 and express it as a linear combination of 81 and 237 (I. e) HCF(81,273=81x+267y for some x and y.

Answer 1

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