Question

if the HCF of 210 and 55 is expressible in the form 210×5+55y then 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

Heya

The HCF of 210 and 55 is 5

210 = 2×3×7×5

55 = 5×11

HCF = 5

5 = 210×5+55y

5-1050 = 55y

-1045/55 = y

-19 = y

Hope it helps