Question

if the hcf of 210 and 55 is expressible in the form 210×5+55y, find y

Answer 1

HCF of 210 and 55 is 5. Given expression for the HCF is 210×5+55y. Therefore 210×5+55y=5 1050+55y=5 55y=-1045 y=-19

Answer 2

Solution:-

Let us first find the HCF of 210 and 55.

Applying Euclid's division lemma on 210 and 55, we get:

210 = 55 × 3 + 45 _____(i)

Since, the remainder 45 ≠ 0. So, we now apply division lemma on the divisor 55 and the remainder 45 to get:

55 = 45 × 1 + 10 _____(ii)

We consider the divisor 45 and the remainder 10 and apply division lemma to get:

45 = 4 × 10 + 5 _____(iii)

We consider the divisor 10 and remainder 5 and apply division lemma to get:

10 = 5 × 2 + 0 ______(iv)

We observe that the remainder at this stage is zero. So, the last divisor i.e. 5 is the HCF of 210 and 55.

Hence, 5 = 210 × 5 + 55y

55y = 5 - 210 × 5 = 5 - 1050

55y = - 1045

y = -1045/55

y = -19