familiar with euclid gcd algorithm ?

not gcd

Ahh gcd is just another name for HCF

yes

Log in to add a comment

familiar with euclid gcd algorithm ?

not gcd

Ahh gcd is just another name for HCF

yes

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

210 = 55 * 3 + 45

55 = 45 * 1 + 10

45 = 10 * 4 + 5

10 = 5 * 2

hence 5 is the HCF.

5 = 45 - 10 * 4**substitute the values of 10 and 45 from above equations.**

= 210 - 55 * 3 - (55 - 45 ) 4

= 210 - 55 * 3 - 55 * 4 + 45 * 4

= 210 - 55 * 7 + (210 - 55 * 3) * 4

= 210 * 5 - 55 * 19

*so x = 5 and y = -19*

============================

*another way : to find an infinite number of solutions.*

210 x + 55 y = 5

42 x + 11 y = 1

11 (4 x + y) = 1 + 2 x

y is an odd number as RHS is odd and so LHS has to be odd. let y = 2 z+1. one of x and y is negative and the other is positive.

11 (4x + 2z+1) = 1 + 2x

42 x + 22 z + 10 = 0

21 x + 11 z + 5 = 0 --- (1)

11 (x+z) = - 5 (2 x + 1)

x+z is a multiple of -5 and perhaps (2x+1) /11

2 x +1 is an odd multiple of 11.

so let 2x + 1 = (2 v + 1) 11

2 x = 22 v + 10

x = 11 v + 5

then by use of (1),

11 z = - 5 - 21 * (11 v + 5) = - 5 - 231 v - 105 = - 231 v - 110

z = - 21 v - 10

y = - 42 v - 19

* 5 = HCF = 210 * (11 v + 5) - 55 * (42 v + 19)*

* *

* x = 11 v + 5*

* y = - (42 v + 19)*

* v x y 210 * x + 55 * y*

0 5 -19 5

-1 -6 23 5

1 16 -61 5

55 = 45 * 1 + 10

45 = 10 * 4 + 5

10 = 5 * 2

hence 5 is the HCF.

5 = 45 - 10 * 4

= 210 - 55 * 3 - (55 - 45 ) 4

= 210 - 55 * 3 - 55 * 4 + 45 * 4

= 210 - 55 * 7 + (210 - 55 * 3) * 4

= 210 * 5 - 55 * 19

============================

210 x + 55 y = 5

42 x + 11 y = 1

11 (4 x + y) = 1 + 2 x

y is an odd number as RHS is odd and so LHS has to be odd. let y = 2 z+1. one of x and y is negative and the other is positive.

11 (4x + 2z+1) = 1 + 2x

42 x + 22 z + 10 = 0

21 x + 11 z + 5 = 0 --- (1)

11 (x+z) = - 5 (2 x + 1)

x+z is a multiple of -5 and perhaps (2x+1) /11

2 x +1 is an odd multiple of 11.

so let 2x + 1 = (2 v + 1) 11

2 x = 22 v + 10

x = 11 v + 5

then by use of (1),

11 z = - 5 - 21 * (11 v + 5) = - 5 - 231 v - 105 = - 231 v - 110

z = - 21 v - 10

y = - 42 v - 19

0 5 -19 5

-1 -6 23 5

1 16 -61 5