Question

if the HCF 210 and 55 is expressible in the form of to 210 * 5 + 55 x then find the value of x ​

Answer 1

$<b><marquee behavior= "alternate">Heya!! </marquee>$

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

$<marquee behavior=move bgcolor=pink> <h1> Follow me ❤</h1></marquee>$

Answer 2

Answer:

Let us first find the HCFof 210 and 55.

Applying Euclids division lemma on 210 and 55, we get

210=55×3+45  

Since the remainder 45=0. So, we now apply division lemma on the divisor 55 and the remainder 45 to get

55=45×1+10  

We consider the divisor 45 and the remainder 10 and apply division lemma to get

45=4×10+5  

We consider the divisor 10 and the remainder 5 and apply division lemma  to get 

10=5×2+0 

We observe that the remainder at this stage is zero. So, that last divisor i.e. 

5 is the HCF of 210 and 55.

∴5=210×5+55y

⇒55y=5−210×5=5−1050

⇒55y=−1045

⇒y=55−1045=−19