HCF of 100, 105, 125, 128 using Euclid's algorithm

Below detailed show work will make you learn how to find HCF of 100,105,125,128 using the Euclidean division algorithm. So, follow the step by step explanation & check the answer for HCF(100,105,125,128).

Here 105 is greater than 100

Now, consider the largest number as 'a' from the given number ie., 105 and 100 satisfy Euclid's division lemma statement a = bq + r where 0 ≤ r < b

Step 1: Since 105 > 100, we apply the division lemma to 105 and 100, to get

105 = 100 x 1 + 5

Step 2: Since the reminder 100 ≠ 0, we apply division lemma to 5 and 100, to get

100 = 5 x 20 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 5, the HCF of 100 and 105 is 5

Notice that 5 = HCF(100,5) = HCF(105,100) .

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

Here 125 is greater than 5

Now, consider the largest number as 'a' from the given number ie., 125 and 5 satisfy Euclid's division lemma statement a = bq + r where 0 ≤ r < b

Step 1: Since 125 > 5, we apply the division lemma to 125 and 5, to get

125 = 5 x 25 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 5, the HCF of 5 and 125 is 5

Notice that 5 = HCF(125,5) .

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

Here 128 is greater than 5

Now, consider the largest number as 'a' from the given number ie., 128 and 5 satisfy Euclid's division lemma statement a = bq + r where 0 ≤ r < b

Step 1: Since 128 > 5, we apply the division lemma to 128 and 5, to get

128 = 5 x 25 + 3

Step 2: Since the reminder 5 ≠ 0, we apply division lemma to 3 and 5, to get

5 = 3 x 1 + 2

Step 3: 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 5 and 128 is 1

Notice that 1 = HCF(2,1) = HCF(3,2) = HCF(5,3) = HCF(128,5) .

Therefore, HCF of 100,105,125,128 using Euclid's division lemma is 1.