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8. If d is the HCF of 468 and 222, find the value of integers x and y which satisfy d-468r+222 |
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Answer» HCF of 468 and 222 468 = (222 x 2) + 24222 = (24 x 9) + 624 = (6 x 4) + 0 Therefore, HCF = 6 6 = 222 - (24 x 9)= 222 - {(468 – 222 x 2) x 9 [where 468 = 222 x 2 + 24]= 222 - {468 x 9 – 222 x 2 x 9}= 222 - (468 x 9) + (222 x 18)= 222 + (222 x 18) - (468 x 9)= 222[1 + 18] – 468 x 9= 222 x 19 – 468 x 9= 468 x -9 + 222 x 19 Hence, HCF of 468 and 222 in the form of 468x + 222y is 468 x -9 + 222 x 19. |
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