# Find the HCFof 65 and 117 and express it in the 65m+117n

i will give u the answer

• -3

• -5

HCF of 117 and 65,

117=65*1+52

65=52*1+13

52=13*4+0

Therefore HCF of 65 and 117 is 13

• -4

HCF 13 can be expressed in this form

65*5+117(-1)=13

=> 130-117=13

So value of m is 5 and value of n is -1

• -13

Sorry a big mistake,

HCF 13 can be expressed in the following form,

65(2)+117(-1)=13

=>130-117=13

So value of m is 2 and value of n is -1

• -5

HCF of 117 and 65,

117=65*1+52

65=52*1+13

52=13*4+0

Therefore HCF of 65 and 117 is 13

HCF 13 can be expressed in the following form,

65(2)+117(-1)=13

=>130-117=13

So value of m is 2 and value of n is -1

• 0

See this also in steps :

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 obtained
13 = 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 65m + 117 n, where m = 2 and n = –1

Cheers!!!
• 72

give thumbs up

• -7
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 obtained
13 = 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 65m + 117 n, where m = 2 and n = –1

Cheers!!!
• 2

HCF of 117 and 65,

117=65*1+52

65=52*1+13

52=13*4+0

Here remainder becomes 0.

Therefore HCF of 65 and 117 is 13

HCF 13 can be expressed in the following form,

65(2)+117(-1)=13

=>130-117=13

So value of m is 2 and value of n is -1

• 30

Ishan..u have copied from me

• -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 65m + 117 n, where m = 2 and n = –1

• 20
I am not getting how 2 has came as m
• 3

By Euclid's division algorithm

117 = 65x1 + 52.

65 = 52x1 + 13

52 = 13x4 + 0

Therefore 13 is the HCF (65, 117).

Now work backwards:

13 = 65 + 52x(-1)

13 = 65 + [117 + 65x(-1)]x(-1)

13 = 65x(2) + 117x(-1).

∴ m = 2 and n = -1.

• 7
Aadhav...m will not be equal to 5!!
• -6
I cannot understand this question till now
• -3
Sandhya
• -3
Jo sab ne diya hai vo right hai...ok
• -2
m is 2 and n is -1
• 0
The HCF(65,117)=13
13=65-52. .....(1)

52=117-65
The from eqn 1
13=65-(117-65)
13=65-117+65
13=65?2+117(-1)

So the value of M&N is 2,-1
• 0
By Euclid's division algorithm 117 = 65x1 + 52. 65 = 52x1 + 13 52 = 13x4 + 0 Therefore 13 is the HCF (65, 117). Now work backwards: 13 = 65 + 52x(-1) 13 = 65 + [117 + 65x(-1)]x(-1) 13 = 65x(2) + 117x(-1). ∴ m = 2 and n = -1.
• 0
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