find the HCF of 52 and 117 and also find the value of x and y if it is expressed in the form of 52x + 117 y
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Heya,,,,
Ur answer is here
_________________________
we Find HCF of (52 , 117 )
using Euclid's Algorithm formula-
On applying Euclid's Algorithm
dividing 117 by 52, we get quotient = 2 and remainder is 13
=> 117 = 52 × 2 + 13
Again on applying Euclid's Algorithm on dividing 52 by 13 , we get
quotient =4 and remainder is 0.
In this step the remainder is zero.
Thus, the division I.e. 4 in this step is the H.C.F of the given Numbers
The HCF of 52 and 117 is 4.
Now,
=> 13 = ( 117 × 1 ) - ( 52 × 2 )
=> 13 = - (52 × 2) + (117 × 1 )
=> 13 = 52x + 117y
=> x = (-2) , y = 1
THEREFORE , HCF OF 52 and 117 is of the form 52x + 117 y ,
where , x = -2 and y = 1
_______________________________
∆Hope it's helps you.
@isharoy688
Thank you ☺☺
Ur answer is here
_________________________
we Find HCF of (52 , 117 )
using Euclid's Algorithm formula-
On applying Euclid's Algorithm
dividing 117 by 52, we get quotient = 2 and remainder is 13
=> 117 = 52 × 2 + 13
Again on applying Euclid's Algorithm on dividing 52 by 13 , we get
quotient =4 and remainder is 0.
In this step the remainder is zero.
Thus, the division I.e. 4 in this step is the H.C.F of the given Numbers
The HCF of 52 and 117 is 4.
Now,
=> 13 = ( 117 × 1 ) - ( 52 × 2 )
=> 13 = - (52 × 2) + (117 × 1 )
=> 13 = 52x + 117y
=> x = (-2) , y = 1
THEREFORE , HCF OF 52 and 117 is of the form 52x + 117 y ,
where , x = -2 and y = 1
_______________________________
∆Hope it's helps you.
@isharoy688
Thank you ☺☺
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