if HCF of 210 and 55 is expressible in the form 210×5-55y,then find the value of x.
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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.
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