If the hcf of 210 and 55 is expressible in the form 210*5+55y then fin

1.	If the hcf of 210 and 55 is expressible in the form 210*5+55y then find y
Answer» Applying Euclid division lemna on 210 and 55, we get210 = 55 × 3 + 45☺️☺️55 = 45 × 1 + 10 ☺️45 = 4 × 10 + 5 ☺️10 = 5 × 2 + 0 ☺️☺️We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55.☺️∴ 5 = 210 × 5 + 55y☺️⇒ 55y = 5 - 1050 = -1045☺️☺️∴ y = -19

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