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

Hi!

Here is the proof to your query.

Among 81 and 237; 237 > 81

Since 237 > 81, we apply the division lemma to 237 and 81 to obtain

237 = 81 × 2 + 75 … Step 1

Since remainder 75 ≠ 0, we apply the division lemma to 81 and 75 to obtain

81 = 75 × 1 + 6 … Step 2

Since remainder 6 ≠ 0, we apply the division lemma to 75 and 6 to obtain

75 = 6 × 12 + 3 … Step 3

Since remainder 3 ≠ 0, we apply the division lemma to 6 and 3 to obtain

6 = 3 × 2 + 0 … Step 4

In this step the remainder is zero. Thus, 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

From Step 3:

3 = 75 – 6 × 12 … Step 5

From Step 2:

6 = 81 – 75 × 1

Thus, from Step 5, it is obtained

3 = 75 – (81 – 75 × 1) × 12

⇒ 3 = 75 – (81× 12 – 75 × 12)

⇒ 3 = 75 × 13 – 81× 12 … Step 6

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)

∴HCF, 3 can be expressed as linear combination of 81 and 237 as 3 = 81x + 237y, where x and y are not unique.

Cheers!

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