Question

HCF of 156, 221, 390 by Division method. Explain with process.​

Answer 1

Answer:

HCF of 103, 156, 221 is 1 the largest number which exactly divides all the numbers i.e. where the remainder is zero. Let us get into the working of this example.

Consider we have numbers 103, 156, 221 and we need to find the HCF of these numbers. To do so, we need to choose the largest integer first and then as per Euclid's Division Lemma a = bq + r where 0 ≤ r ≤ b

steps

Step 1: Since 156 > 103, we apply the division lemma to 156 and 103, to get

156 = 103 x 1 + 53

Step 2: Since the reminder 103 ≠ 0, we apply division lemma to 53 and 103, to get

103 = 53 x 1 + 50

Step 3: We consider the new divisor 53 and the new remainder 50, and apply the division lemma to get

53 = 50 x 1 + 3

We consider the new divisor 50 and the new remainder 3,and apply the division lemma to get

50 = 3 x 16 + 2

We consider the new divisor 3 and the new remainder 2,and apply the division lemma to get

3 = 2 x 1 + 1

We consider the new divisor 2 and the new remainder 1,and apply the division lemma to get

2 = 1 x 2 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 1, the HCF of 103 and 156 is 1

Notice that 1 = HCF(2,1) = HCF(3,2) = HCF(50,3) = HCF(53,50) = HCF(103,53) = HCF(156,103) .

We can take hcf of as 1st numbers and next number as another number to apply in Euclidean lemma

Step 1: Since 221 > 1, we apply the division lemma to 221 and 1, to get

221 = 1 x 221 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 1, the HCF of 1 and 221 is 1

Notice that 1 = HCF(221,1) .

Answer 2

Step-by-step explanation:

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