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 464, 536 i.e. 8 the largest integer that leaves a remainder zero for all numbers.
HCF of 464, 536 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 464, 536 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 464, 536 is 8.
HCF(464, 536) = 8
Highest common factor or Highest common divisor (hcd) can be calculated by Euclid's algotithm.
Highest common factor (HCF) of 464, 536 is 8.
Step 1: Since 536 > 464, we apply the division lemma to 536 and 464, to get
536 = 464 x 1 + 72
Step 2: Since the reminder 464 ≠ 0, we apply division lemma to 72 and 464, to get
464 = 72 x 6 + 32
Step 3: We consider the new divisor 72 and the new remainder 32, and apply the division lemma to get
72 = 32 x 2 + 8
We consider the new divisor 32 and the new remainder 8, and apply the division lemma to get
32 = 8 x 4 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 8, the HCF of 464 and 536 is 8
Notice that 8 = HCF(32,8) = HCF(72,32) = HCF(464,72) = HCF(536,464) .
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 464, 536?
Answer: HCF of 464, 536 is 8 the largest number that divides all the numbers leaving a remainder zero.
3. How to find HCF of 464, 536 using Euclid's Algorithm?
Answer: For arbitrary numbers 464, 536 apply Euclid’s Division Lemma in succession until you obtain a remainder zero. HCF is the remainder in the last but one step.