Question

Find HCF of 81 and 237 and express it as a linear combination of 81 and 237 i.e., HCF (81,237)= 81x + 237y for some x and y. Please explain.

Answer 1

Between 81 and 237; 237 is greater than 81

Division lemma of 237 and 81:

Step 1: 237 = 81 × 2 + 75

Step 2: Since remainder 75 ≠ 0, division lemma is applied to 81 and 75 to get 81 = 75 × 1 + 6

Step 3: Since remainder 6 ≠ 0, division lemma is applied to 75 and 6 to get 75 = 6 × 12 + 3

Step 4: Since remainder 3 ≠ 0, division lemma is applied to 6 and 3 to get 6 = 3 × 2 + 0

The remainder is zero in step 4.

Therefore, the divisor i.e. 3 in this step is the H.C.F. of the given numbers.

The H.C.F. of 237 and 81 is 3

Step 5: From Step 3: 3 = 75 – 6 × 12 -----

From Step 2: 6 = 81 – 75 × 1

Thus, from Step 5, we, get 3 = 75 – (81 – 75 × 1) × 12

⇒ 3 = 75 – (81× 12 – 75 × 12)

Step 6 ⇒ 3 = 75 × 13 – 81× 12

From Step 1, 75 = 237 – 81 × 2

Thus, from Step 6;

3 = (237 – 81 × 2) × 13 – 81× 12
⇒ 3 = (237 × 13 – 81 × 26) – 81× 12
⇒ 3 = 237 × 13 – 81 × 38
⇒ H.C.F. of 237 and 81 = 237 × 13 + 81 × (–38)

237 × 13 + 81 × (–38) is the representation of H.C.F. of 237 and 81 as linear combination of 237 and 81. 237x + 81y where x = 13 & y = (-38) Hope u got it