Question

If the HCF (420,130) = 420x +130y, then the values of x and y satisfying the above relations are​

Answer 1

Answer:

Answer:

We'll follow the Euclid Algorithm to solve this problem,

420 = 3*130 + 30...(1)

Now,

130 = 4*30 + 10 ...(2)

30 = 3*10+0...(3)

Hence the HCF of both these numbers will be 10.

From equation 2 :

HCF (420,130) = 10 = (130-4*30)

and, 30 = 420-(3*130)

So,

10 = (130-4*(420-3*130)) = 13*130 + (-4)*420...(4)

And hence we've shown that the GCD can be shown as a linear combination

To prove that it's not unique

Let's add and subtract the number

(420)*(130)*m

to equation 4

We get

10 = 13*130 + (-4)*420 + (420m)*130 - (130m)*420

=(13+420m)*130 + (-4-130m)*420

So, we can clearly see that on putting in different values of m as an integers we can get different ways of expressing the HCF as a linear combination of both the number