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If d =HCF of 45 and 27 find x and y such that d=27x+45y

Posted by Muskan Yadav 5 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Hdjf Kumari 5 years, 7 months ago

45=27×1+18 27=18×1+9 18=9×2+0 HCF=divisor=9 d=HCF of 45 and 27 d=9 d=27-18 d=27-(45-27) d=27×2-45 d=27x+45y Here,x=2 and y=-1

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