if D is the HCF of 468 and 222 find the value of integers X and Y which satisfy the equal to 468 X + 222 y
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HCF of 468 and 222
468 = (222 x 2) + 24
222 = (24 x 9) + 6
24 = (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.
hope rhis will help you...
468 = (222 x 2) + 24
222 = (24 x 9) + 6
24 = (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.
hope rhis will help you...
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