Question

if the HCF of 210 and 55 is expressible in the form 210 x 5 + 55y then find y.

Answer 1

Euclid division lemma:-
a=bq+r

210 = 55(3)+ 45
55 = 45(1) + 10
45 = 4(10)+ 5
10 = 5(2)+ 0

Hcf is 5

5=210×5+55y
-55y=210(5)-5
-55y=1050-5
-55y=1045
y=1045/-55
y= -19

Hope it helps..

Answer 2

Answer:

The value of y is 19.

Step-by-step explanation:

SOLUTION :  

Given : Two numbers 210 & 55  

HCF(210,55) = 210 × 5 - 55y………(1)

By using Euclid division lemma, a = bq+ r,  

210 = 55 × 3 + 45

55 = 45 × 1 + 10  

45 = 4 × 10 + 5  

10 = 5 × 2 + 0  

Here Remainder is zero and last divisor is 5.  

Hence HCF of 210 & 55 is 5.

On putting HCF(210,55) = 5 in eq 1,

HCF(210,55) = 210 × 5 - 55y

5 = 210 × 5 - 55y

5 = 1050 - 55y

55y = 1050 - 5

55y = 1045

y = 1045/55

y = 19

Hence,the value of y is 19.

HOPE THIS ANSWER WILL HELP YOU…..