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 608, 264 i.e. 8 the largest integer that leaves a remainder zero for all numbers.
HCF of 608, 264 is 8 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 608, 264 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 608, 264 is 8.
HCF(608, 264) = 8
Highest common factor or Highest common divisor (hcd) can be calculated by Euclid's algotithm.
Highest common factor (HCF) of 608, 264 is 8.
Step 1: Since 608 > 264, we apply the division lemma to 608 and 264, to get
608 = 264 x 2 + 80
Step 2: Since the reminder 264 ≠ 0, we apply division lemma to 80 and 264, to get
264 = 80 x 3 + 24
Step 3: We consider the new divisor 80 and the new remainder 24, and apply the division lemma to get
80 = 24 x 3 + 8
We consider the new divisor 24 and the new remainder 8, and apply the division lemma to get
24 = 8 x 3 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 8, the HCF of 608 and 264 is 8
Notice that 8 = HCF(24,8) = HCF(80,24) = HCF(264,80) = HCF(608,264) .
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 608, 264?
Answer: HCF of 608, 264 is 8 the largest number that divides all the numbers leaving a remainder zero.
3. How to find HCF of 608, 264 using Euclid's Algorithm?
Answer: For arbitrary numbers 608, 264 apply Euclid’s Division Lemma in succession until you obtain a remainder zero. HCF is the remainder in the last but one step.