if d is the HCF of 45&27 find X&y satisfying d=27x+45y
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If d is the HCF of 45 and 27 find x and y satisfying d= 27x+45y
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Applying Euclid's division lemma to 27 and 45, we get
45=27×1+18 ...(1)
27=18×1+9 ...(2)
18=9×2+0 ...(3)
Since the remainder is zero, therefore, last divisor 9 is the HCF of 27 and 45.
From (2), we get
9=27−18×1
=27−(45−27×1)×1 [using(1)]
=27−45×1+27×1×1
=27−45+27
=54−45
⇒9=27×2−45×1 ...(4)
Comparing (4) with d=27x+45y, we get
d=9,x=2 and y=−1
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