Question

If the H.C.F of 210 and 55 is expressible in the form 210x5+55y, find y.​

Answer 1

Answer:

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Answer 2

Let us first find the HCFof 210 and 55.

Applying Euclids division lemma on 210 and 55, we get

210=55×3+45

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

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

45=4×10+5

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

10=5×2+0

We observe that the remainder at this stage is zero. So, that last divisor i.e.

5 is the HCF of 210 and 55.

∴5=210×5+55y

⇒55y=5−210×5=5−1050

⇒55y=−1045

⇒y=

55

−1045

=−19