Question

10.If d is the HCF of 56 and 72 find x y satisfying d=56x+72y. Also show that x and y are not unique. ​

Answer 1

Question

If D is the HCF of 56 and 72,find x and y satisfying D = 56x + 72y. Also show that x and y are not unique

Solution

Given,

D is the HCF of 56 and 72

According to Euclid's Division Lemma,

$\tt{72 = 56 \times 1 + 16 - - - - - - (1)}$

As r ≠ 0,

$\longrightarrow \: \tt{56 = 16 \times 3 + 8 - - - - - (2)}$

As r ≠ 0,

$\longrightarrow \: \tt{16 = 8 \times 2 + 0 - - - - - (3)}$

Therefore,8 is the HCF of 56 and 72

Now,

$\boxed{ \boxed{ \tt{8 = 56x + 72y}}} - - - - - (4)$

From Equation (2),

$\tt{8 = 56 - 16 \times 3} \\ \\ \implies \: \tt{8 = 56 - (72 - 56 \times 1) \times 3} \\ \\ \implies \: \sf{8 = 56(4) + 72( - 3)}$

On comparing with equation (4),we infer :

$\tt{x = 4 \: and \: y = - 3}$

Also,

$\tt{8 = 56 \times 4 + 72 \times - 3} \\ \\ \implies \: \tt{8 = 56 \times 4 + 72 \times ( - 3) + (56 \times 72) - (56 \times 72)} \\ \\ \implies \: \tt{8 = 56(4 - 72) +72 ( - 3 + 56)} \\ \\ \implies \: \tt{8 = 56( - 68) + 72(53)}$

On comparing with (4),we infer :

$\tt{x = - 68 \: and \: y = 53}$

Hence, x and y aren't unique for D = 56x + 72y