If the HCF of 210 and 55 is expressed in the form 210×5+55y, find y

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Solution

Let us first find the HCF of 210 and 55.
210 = 5 x 2 x 3 x 7
55= 5 x 11
HCF = 5
∴ 5 = 210 × 5 + 55y

⇒ 55y = 5 - 1050 = -1045

∴ y = -19

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