Question

If d is the HCF of 63 and 81, find the values of x and y satisfying 'd'=63x+81y where x and y are unique

Answer 1

By using : 
a = bq + r  
  
81 =  63 * 1 + 18
63 =  18 * 3 + 9
18 =    9 * 2 + 0
 H.C.F = 9

= 63 - ( 18 * 3 )
= 63 - 3*18
= 63 - 3 [ 81 - ( 63 * 1 ) ]
= 63 - 3 [ 81 - 63 ]
= 63 - 81(3) + 63(3)
= 63(4) + 81(-3)

So, In 63x + 81y
 x = 4 and y = -3   Answer

Answer 2

Answer:

Concept:

Euclid's Division Lemma (lemma is like a theorem) says that given two positive integers a and b, there exist unique integers q and r such that

a = bq + r, 0≤ r <b

Step-by-step explanation:

Given:

d is the HCF of 63 and 81

Where, d = 63X + 81Y

Find:

Values of X and Y

Solution:

so by substituting the given values and equation,

$\begin{aligned}&81=63 * 1+18 \\&63=18 * 3+9 \\&18=9 * 2+0 \\&\text { H.C.F }=9 \\&=63-(18 * 3) \\&=63-3^{*} 18 \\&=63-3[81-(63 * 1)] \\&=63-3[81-63] \\&=63-81(3)+63(3) \\&=63(4)+81(-3) \\&\text { So, In } 63 x+81 y \\&x=4 \text { and } y=-3 \text { }\end{aligned}$

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