find the value of h.c.f of 65 & 117 is expressible is form of 65n & 117
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Question : If the HCF of 65 and 117 is of the form (65n-117) then find the value of n
In Euclid division lemma, We try to express one number as a multiple of number and we continue till we get a smallest number which leaves 0 as remainder.
Given, Two numbers 79 , 97 .
First we determine which of them is the biggest.
So 117 > 65
We now try to express this as
117 = 65c + d for something and continue till d = 0
Now, 117 = 65*1 + 52
We see that, Remainder is not 0 .
So repeat the same with 52 , 65( 65 > 52 )
65 = 52 * 1 + 13
Still the remainder, is not zero.
Again repeating with ( 13 , 52 )
52 = 13 * 4 + 0
Therefore, The required HCF is 13
From the above equations,
13 = 65 - ( 52 )
13 = 65 - ( 117- 65)
13 = 65*2 - 117
Given that ,HCF ( 65 , 117 ) = 65m - 117
Now, 65 * 2 - 117 = 65n- 117
So, n = 2
Hope helped!
Question : If the HCF of 65 and 117 is of the form (65n-117) then find the value of n
In Euclid division lemma, We try to express one number as a multiple of number and we continue till we get a smallest number which leaves 0 as remainder.
Given, Two numbers 79 , 97 .
First we determine which of them is the biggest.
So 117 > 65
We now try to express this as
117 = 65c + d for something and continue till d = 0
Now, 117 = 65*1 + 52
We see that, Remainder is not 0 .
So repeat the same with 52 , 65( 65 > 52 )
65 = 52 * 1 + 13
Still the remainder, is not zero.
Again repeating with ( 13 , 52 )
52 = 13 * 4 + 0
Therefore, The required HCF is 13
From the above equations,
13 = 65 - ( 52 )
13 = 65 - ( 117- 65)
13 = 65*2 - 117
Given that ,HCF ( 65 , 117 ) = 65m - 117
Now, 65 * 2 - 117 = 65n- 117
So, n = 2
Hope helped!
aadi93:
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