Question

If d is the HCF of 468&222.Find the value of integers x and y which satisfy d=468x+222y.

Answer 1

HCF(468,222)
468= 222×2+24
222= 24×9 +6
24= 6×4+0
=> HCF(468,222) = 6
NOW,
6= 222 - ( 24×9)
= 222- [(468-222×2)×9]
= 222 - 468×9 + 222×2×9
= 468×(-9) + 222 ( 1+ 2×9)
= 468(-9) + 222(19)
On comparing,
X= -9 and Y= 19