Question

If the HCF of 210 and 55 is expressible in the form of 210×5×55×A ,then find the value of (2-A)​

Answer 1

Answer:

y=19

Step-by-step explanation:

Since now the remainder is zero, therefore divisor at this stage, that is 5, is the HCF of 210 and 55. Comparing equation (vi) with the given equation: 5=210×5+55y, we get, y=−19. This is the required solution.

Answer 2

Answer:

Thus 2-A=2-(-19)=21

Step-by-step explanation:

First applying division lemma on 210 and 55

210=3*55+45

Now applying division lemma on 55 and 45

55=45*1+10

Now applying division lemma on 45 and 10

45=10*4+5

Now applying division lemma on 10 and 5

10=2*5+0

Here remainder is zero

So HCF(210,55)=5

Given that

HCF=210*5+55*A

5=210*5+55A

55A=5-1050=-1045

A=-1045/5=-19

Thus 2-A=2-(-19)=21