Created By : Jatin Gogia
Reviewed By : Rajasekhar Valipishetty
Last Updated : Apr 06, 2023
HCF Calculator using the Euclid Division Algorithm helps you to find the Highest common factor (HCF) easily for 268, 868 i.e. 4 the largest integer that leaves a remainder zero for all numbers.
HCF of 268, 868 is 4 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 268, 868 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
Highest common factor (HCF) of 268, 868 is 4.
HCF(268, 868) = 4
Highest common factor or Highest common divisor (hcd) can be calculated by Euclid's algotithm.
Highest common factor (HCF) of 268, 868 is 4.
Step 1: Since 868 > 268, we apply the division lemma to 868 and 268, to get
868 = 268 x 3 + 64
Step 2: Since the reminder 268 ≠ 0, we apply division lemma to 64 and 268, to get
268 = 64 x 4 + 12
Step 3: We consider the new divisor 64 and the new remainder 12, and apply the division lemma to get
64 = 12 x 5 + 4
We consider the new divisor 12 and the new remainder 4, and apply the division lemma to get
12 = 4 x 3 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 268 and 868 is 4
Notice that 4 = HCF(12,4) = HCF(64,12) = HCF(268,64) = HCF(868,268) .
Here are some samples of HCF using Euclid's Algorithm calculations.
1. What is the Euclid division algorithm?
Answer: Euclid's Division Algorithm is a technique to compute the Highest Common Factor (HCF) of given positive integers.
2. what is the HCF of 268, 868?
Answer: HCF of 268, 868 is 4 the largest number that divides all the numbers leaving a remainder zero.
3. How to find HCF of 268, 868 using Euclid's Algorithm?
Answer: For arbitrary numbers 268, 868 apply Euclid’s Division Lemma in succession until you obtain a remainder zero. HCF is the remainder in the last but one step.