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CBSE Class 10 Answered

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​

Asked by adnankhan7860098 | 21 Sep, 2021, 06:21: PM
Expert Answer
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
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