Question

If the H.C.F of 210 and 55 is expressible in the form of 210×5+55y, find the value of y.​

Answer 1

Answer:

Since now the remainder is zero, therefore divisor at this stage, that is 5, is the HCF of 210 and 55. Comparing equation (vi) with the given equation: 5=210×5+55y, we get, y=−19. This is the required solution

Answer 2

Answer:

Since now the remainder is zero, therefore divisor at this stage, that is 5, is the HCF of 210 and 55. Comparing equation (vi) with the given equation: 5=210×5+55y, we get, y=−19. This is the required solution

Step-by-step explanation:

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