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IF d IS H.C.F OF 56 AND 72 FIND x,y SATISFYING d=56x+72y. Also show that x and y are not unique
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).
According to Euclid division Lemma72=56*1+1656=16*3+816=8*2+0....... Therefore 8 is the HCF. Now,,,, 8=56-(16*3)..... In this step we have expressed 8 in form of 56 and 16, we have to express it in the form of 56 and 72. So, Now by splliting 16 we get,. 8=56-(72-56*1)(3) 8 =56-72*3+56*3....... As we have another 56 we group them 8 =56*4+72(-3). Hence X and Y are 4,3 respectively .. Therefore X and Y are not equal
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