Question

find the HCF of 56,72 using Euclid's division method​

Answer 1

$56 = {2}^{3} \times 7$

$72 = {2}^{3} \times {3}^{2}$

$hcf = {2}^{3} = 8$

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Answer 2

By Euclid division Lemma,

72 = 56 × 1 + 16 --------> (i)

56 = 16 × 3 + 8 ----------> (ii)

16 = 8 × 2 + 0

So, 8 is the GCF or HCF of 56 and 72.

From (ii) 8 = 56 - (16 × 3)

⇒ 8 = 56 - [( 72 - 56) × 3] from (i)

⇒ 8 = 56 - 72(3) + 56(3)

⇒ 8 = 56(4) - 72(3)

⇒ 8 = 56(4) + 72(-3)

⇒ d = 8 = 56(4) + 72(-3) = 56x + 72y

one set of values for x and y are 4 , -3 respectively.

If we rewrite:

8 = [56 × 4 + (-3) × 72 ] - 56 × 72 + 56 × 72

⇒ 8 = [56 × 4 - 56 × 72] + [(-3) × 72 + 56 × 72]

⇒ 8 = [56(4 - 72)] + [72(-3+56)]

⇒ 8 = 56(-68) + 72(52)

Now, we get a new set of values for x and y as -68 and 52 respectively.

Therefore, x and y are not unique

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