Question

if HCF (210,55) is expressible in the form 210×5-55y find y​

Answer 1

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.