If the HCF of 210 and 55 is expressible in the form 210×5+55y, find y.
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=−104555=−19