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 594, 282 i.e. 6 the largest integer that leaves a remainder zero for all numbers.
HCF of 594, 282 is 6 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 594, 282 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 594, 282 is 6.
HCF(594, 282) = 6
Highest common factor or Highest common divisor (hcd) can be calculated by Euclid's algotithm.
Highest common factor (HCF) of 594, 282 is 6.
Step 1: Since 594 > 282, we apply the division lemma to 594 and 282, to get
594 = 282 x 2 + 30
Step 2: Since the reminder 282 ≠ 0, we apply division lemma to 30 and 282, to get
282 = 30 x 9 + 12
Step 3: We consider the new divisor 30 and the new remainder 12, and apply the division lemma to get
30 = 12 x 2 + 6
We consider the new divisor 12 and the new remainder 6, and apply the division lemma to get
12 = 6 x 2 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 594 and 282 is 6
Notice that 6 = HCF(12,6) = HCF(30,12) = HCF(282,30) = HCF(594,282) .
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 594, 282?
Answer: HCF of 594, 282 is 6 the largest number that divides all the numbers leaving a remainder zero.
3. How to find HCF of 594, 282 using Euclid's Algorithm?
Answer: For arbitrary numbers 594, 282 apply Euclid’s Division Lemma in succession until you obtain a remainder zero. HCF is the remainder in the last but one step.