Question

Heyy guys... plzz help me...

If D is the HCF of 468 and 222 find the value of integers X and Y which satisfy D = 468 X + 222Y

Answer 1

Heya......¡!!here you go.....

Answer 2

Heya friend,


Here is your answer,

Given : d is the HCF of 468 and 222 and d = 468x + 222y

Solution :
Let,
=> 468 = (222 × 2) + 24 .....................(1)

=> 222 = (24 × 9) + 6 ....................... (2)

=> 24 = (6 × 4) +0

Since,
H.C.F. = d = 6

From (2) , we have

6 = 222 - (24 x 9) .....................(3)

Putting value of 24 from (1) in (3)

=> 6 = 222 - [ {468 - ( 222 x 2)} x 9 ]

=> d = 222 - [ (468 x 9 ) - (222 x 2 x 9) ]

=> d = (222 x 1) - [ (468 x 9) - ( 222 x 18)]

=> d = (222 x 1) - (468 x 9 ) + (222 x 18 )

=> d = 222 x (1+18) - (468 x 9) (just took 222 as common)

=> d = 222 x 19 + 468 x (-9)

d = 468 x (-9) + 222 x 19

Now, comparing with the given equation,
d = 468x +222y

Therefore,
x = -9 and y = 19 (ans)


Hope it helps you.
Thank you.