If a is the HCF of 56 and 72.Find the value of x and y satisfying a=56x+12y.show that x and y are not unique.
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Hi ...here is your answer...
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.. as we know that .....
According to Euclid division
Lemma
72=56*1+1656
=16*3+816
=8*2+0 T
herefore 8 is the HCF.
8=56-(16*3)
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 Therefore X and Y are uniquel
.
.. as we know that .....
According to Euclid division
Lemma
72=56*1+1656
=16*3+816
=8*2+0 T
herefore 8 is the HCF.
8=56-(16*3)
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 Therefore X and Y are uniquel
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