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 515, 140 i.e. 5 the largest integer that leaves a remainder zero for all numbers.
HCF of 515, 140 is 5 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 515, 140 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 515, 140 is 5.
HCF(515, 140) = 5
Highest common factor or Highest common divisor (hcd) can be calculated by Euclid's algotithm.
Highest common factor (HCF) of 515, 140 is 5.
Step 1: Since 515 > 140, we apply the division lemma to 515 and 140, to get
515 = 140 x 3 + 95
Step 2: Since the reminder 140 ≠ 0, we apply division lemma to 95 and 140, to get
140 = 95 x 1 + 45
Step 3: We consider the new divisor 95 and the new remainder 45, and apply the division lemma to get
95 = 45 x 2 + 5
We consider the new divisor 45 and the new remainder 5, and apply the division lemma to get
45 = 5 x 9 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 5, the HCF of 515 and 140 is 5
Notice that 5 = HCF(45,5) = HCF(95,45) = HCF(140,95) = HCF(515,140) .
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 515, 140?
Answer: HCF of 515, 140 is 5 the largest number that divides all the numbers leaving a remainder zero.
3. How to find HCF of 515, 140 using Euclid's Algorithm?
Answer: For arbitrary numbers 515, 140 apply Euclid’s Division Lemma in succession until you obtain a remainder zero. HCF is the remainder in the last but one step.