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Here is the answer to your query.

Among 65 and 117; 2117 > 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 obtained13 = 65 – (117 – 65 � 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*= –1Hope! This will help you.

Cheers!!!

- 72

Among 65 and 117; 2117 > 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 obtained13 = 65 – (117 – 65 �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*= –1Hope! This will help you.

Cheers!!!

- 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 obtained13 = 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- 20