if d is the HCF Of 45 and 27. find x and y satisfying d=27x+45y
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45 = 3×3×5
27 = 3×3×3
hcf of 45 and 27 is 9 since 3s are common factors..
now,
d=9
9 = 27x + 45y
9 = 27(2) + 45(-1)
9 = 54 - 45
9 = 9
thus,
x = 2 , y = -1
27 = 3×3×3
hcf of 45 and 27 is 9 since 3s are common factors..
now,
d=9
9 = 27x + 45y
9 = 27(2) + 45(-1)
9 = 54 - 45
9 = 9
thus,
x = 2 , y = -1
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