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

Using Euclid's Division Lemma we find the HCF of two positive integers.

Answer: If d is HCF of 56 and 72, then x, y satisfying the equation d = 56x + 72y are 4, -3 respectively and are not unique.

Let's show x and y are not unique.

Explanation:

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(53)

Now, we get a new set of values for x and y as -68 and 53 respectively.

Therefore, x and y are not unique.