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When the HCF (420, 130) is expressed as a linear combination of 420 and 130 i.e. HCF (420, 130) = 420x + 130y, the values of x and y satisfying the above relation are:  (a)x= 3, y = 1  (c) x = 4, y = -13  (b) x=-4, y = 13  (d) x = 2, y = 3​
By Euclid's division algorithm, we have
420 = 130*3 + 30
130 = 30*4 + 10
30 = 10*3 + 0
So, the HCF will be 10.
i.e. HCF(420, 130) = 10
Now, 10 = 130 - 30*4
Therefore, 10 = 130 - (420 - 130*3)*4
Therefore, 10 = 130*13 + (-4)*420
As HCF (420, 130) = 420x + 130y
Therefore, x = -4 and y = 13
Answered by Renu Varma | 23 Sep, 2021, 01:15: PM

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