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