Question

If the hcf (210,55) is expressible in the form of 210×5-55y, find y..

Answer 1

Heya !!

Let us frist 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 division i.e., 5 is the HCF of 210 and 55.

∴ 5 = 210×5+55y
= 55y = 5-1050 = -1045

∴ y = -19

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@vaibhav246

Answer 2

Hyyyy


Here is the answer



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


Hope helped