If d is the HCF 56 and 72,find x and y satisfiying d=56x+72y
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According to Euclid`s algorithm,
a=bq+r
so 72=56*1+16 ......(1)
56=16*3+8 .....(2)
16=8*2+0 H.C.F (56,72)= 8
Now 8=56-16*3 ......(2)
8=56- (72-56*1)*3 .....(1)
8=56- (72-56)*3 8=56-3(72-56)
8=56-72*3 +56*3 8=56(1+3) +72(-3)
8=56*4 +72(-3)
Taking x=4 and y=-3 8=56x+72y Also x and y not equal to 1(Proved).
a=bq+r
so 72=56*1+16 ......(1)
56=16*3+8 .....(2)
16=8*2+0 H.C.F (56,72)= 8
Now 8=56-16*3 ......(2)
8=56- (72-56*1)*3 .....(1)
8=56- (72-56)*3 8=56-3(72-56)
8=56-72*3 +56*3 8=56(1+3) +72(-3)
8=56*4 +72(-3)
Taking x=4 and y=-3 8=56x+72y Also x and y not equal to 1(Proved).
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