Question

H.C.F of 728and 216 by E.D.A

Answer 1

Answer:

• The divisor at this stage, ie, 4 is the HCF of 728 and 216.

Step-by-step explanation:

Clearly, 728 > 216

Applying the Euclid's division lemma to 728 and 216, we get

728 = 216 x 3 + 80

Since the remainder 80 ≠ 0, we apply the Euclid's division lemma to divisor 216 and remainder 80 to get

216 = 80 x 2 + 56

We consider the new divisor 80 and remainder 56 and apply the division lemma to get

80 = 56 x 1 + 24

We consider the new divisor 56 and remainder 24 and apply the division lemma to get

56 = 24 x 2 + 8

We consider the new divisor 24 and remainder 8 and apply the division lemma to get

24 = 8 x 3 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 4 is the HCF of 728 and 216.

Answer 2

Answer:

• The divisor at this stage, ie, 4 is the HCF of 728 and 216.

Step-by-step explanation:

Clearly, 728 > 216

Applying the Euclid's division lemma to 728 and 216, we get

728 = 216 x 3 + 80

Since the remainder 80 ≠ 0, we apply the Euclid's division lemma to divisor 216 and remainder 80 to get

216 = 80 x 2 + 56

We consider the new divisor 80 and remainder 56 and apply the division lemma to get

80 = 56 x 1 + 24

We consider the new divisor 56 and remainder 24 and apply the division lemma to get

56 = 24 x 2 + 8

We consider the new divisor 24 and remainder 8 and apply the division lemma to get

24 = 8 x 3 + 0

Now, the remainder at this stage is 0.

So, the divisor at this stage, ie, 4 is the HCF of 728 and 216.