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 610, 544 i.e. 2 the largest integer that leaves a remainder zero for all numbers.
HCF of 610, 544 is 2 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 610, 544 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 610, 544 is 2.
HCF(610, 544) = 2
Highest common factor or Highest common divisor (hcd) can be calculated by Euclid's algotithm.
Highest common factor (HCF) of 610, 544 is 2.
Step 1: Since 610 > 544, we apply the division lemma to 610 and 544, to get
610 = 544 x 1 + 66
Step 2: Since the reminder 544 ≠ 0, we apply division lemma to 66 and 544, to get
544 = 66 x 8 + 16
Step 3: We consider the new divisor 66 and the new remainder 16, and apply the division lemma to get
66 = 16 x 4 + 2
We consider the new divisor 16 and the new remainder 2, and apply the division lemma to get
16 = 2 x 8 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 2, the HCF of 610 and 544 is 2
Notice that 2 = HCF(16,2) = HCF(66,16) = HCF(544,66) = HCF(610,544) .
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 610, 544?
Answer: HCF of 610, 544 is 2 the largest number that divides all the numbers leaving a remainder zero.
3. How to find HCF of 610, 544 using Euclid's Algorithm?
Answer: For arbitrary numbers 610, 544 apply Euclid’s Division Lemma in succession until you obtain a remainder zero. HCF is the remainder in the last but one step.