Find the HCF of 56 and 72 and express it in the form of 56x + 72y.
Answers
Answer:
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Answer:
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.
Make your child
naturally math minded
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.
Make your child
naturally math minded