Find the 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 .

To solve the problem, we need to find the HCF of 81 and 237 and express it as a linear combination of 81 and 237 in the form \(81x + 237y = \text{HCF}(81, 237)\).

**Step 1: Find the HCF using the Euclidean algorithm**
1. Divide 237 by 81 and find the remainder:
237 ÷ 81 = 2 with a remainder of 75.
Therefore, 237 = 81 × 2 + 75.

2. Divide 81 by 75 and find the remainder:
81 ÷ 75 = 1 with a remainder of 6.
Therefore, 81 = 75 × 1 + 6.

3. Divide 75 by 6 and find the remainder:
75 ÷ 6 = 12 with a remainder of 3.
Therefore, 75 = 6 × 12 + 3.

4. Divide 6 by 3 and find the remainder:
6 ÷ 3 = 2 with a remainder of 0.
Therefore, 6 = 3 × 2 + 0.

Since the remainder is now 0, the HCF is the last non-zero remainder, which is 3.

**Step 2: Express the HCF as a linear combination**
We will now work backward to express 3 as a linear combination of 81 and 237.

1. From the step \(75 = 6 × 12 + 3\), rewrite 3:
3 = 75 - 6 × 12.

2. Substitute \(6 = 81 - 75 × 1\) (from \(81 = 75 × 1 + 6\)) into the equation:
3 = 75 - (81 - 75 × 1) × 12.
3 = 75 - 81 × 12 + 75 × 12.
3 = 75 × 13 - 81 × 12.

3. Substitute \(75 = 237 - 81 × 2\) (from \(237 = 81 × 2 + 75\)) into the equation:
3 = (237 - 81 × 2) × 13 - 81 × 12.
3 = 237 × 13 - 81 × 26 - 81 × 12.
3 = 237 × 13 - 81 × 38.

Thus, \(3 = 237 × 13 - 81 × 38\).

**Final Answer**
The HCF of 81 and 237 is 3, and it can be expressed as a linear combination:
HCF(81, 237) = 3 = 81(-38) + 237(13).