Question

If the HCF of 210 and 55 is expressible in the form of 210x5+55y , find y

Answer 1

Let us first find the HCF of 210 and 55.

Applying Euclid division lemna on 210 and 55, we get

210 = 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

Answer 2

HCF (210,55) = 5
SO, 5 = 210x + 55y