1>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)

2>

Among 65 and 117; 117 > 65

Since 117 > 65, we apply the division lemma to 117 and 65 to obtain

117 = 65 × 1 + 52** … Step 1**

Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain

65 = 52 × 1 + 13** … Step 2**

Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain

52 = 4 × 13 + 0** … Step 3**

In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers

The H.C.F. of 65 and 117 is 13

From **Step 2**:

13 = 65 – 52 × 1** … Step 4**

From **Step 1**:

52 = 117 – 65 × 1

Thus, from **Step 4**, it is obtained

13 = 65 – (117 – 65 × 1) × 1

⇒13 = 65 × 2 – 117

⇒13 = 65 × 2 + 117 × (–1)

In the above relationship the H.C.F. of 65 and 117 is of the form 65*m* + 117 *n*, where *m* = 2 and *n* = –1

3>

Let us first find the HCF of 210 and 55.

Applying Euclid division lemna on 210 and 55, we get

210 = 55 × 3 + 45 ....(1)

Since the remainder 45 ≠ 0. So, again applying the Euclid division lemna on 55 and 45, we get

55 = 45 × 1 + 10 .... (2)

Again, considering the divisor 45 and remainder 10 and applying division lemna, we get

45 = 4 × 10 + 5 .... (3)

We now, consider the divisor 10 and remainder 5 and applying division lemna to get

10 = 5 × 2 + 0 .... (4)

We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55.

∴ 5 = 210 × 5 + 55y

⇒ 55y = 5 - 1050 = -1045

⇒ y = -19