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 285, 351 i.e. 3 the largest integer that leaves a remainder zero for all numbers.
HCF of 285, 351 is 3 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 285, 351 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 285, 351 is 3.
HCF(285, 351) = 3
Highest common factor or Highest common divisor (hcd) can be calculated by Euclid's algotithm.
Highest common factor (HCF) of 285, 351 is 3.
Step 1: Since 351 > 285, we apply the division lemma to 351 and 285, to get
351 = 285 x 1 + 66
Step 2: Since the reminder 285 ≠ 0, we apply division lemma to 66 and 285, to get
285 = 66 x 4 + 21
Step 3: We consider the new divisor 66 and the new remainder 21, and apply the division lemma to get
66 = 21 x 3 + 3
We consider the new divisor 21 and the new remainder 3, and apply the division lemma to get
21 = 3 x 7 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 3, the HCF of 285 and 351 is 3
Notice that 3 = HCF(21,3) = HCF(66,21) = HCF(285,66) = HCF(351,285) .
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 285, 351?
Answer: HCF of 285, 351 is 3 the largest number that divides all the numbers leaving a remainder zero.
3. How to find HCF of 285, 351 using Euclid's Algorithm?
Answer: For arbitrary numbers 285, 351 apply Euclid’s Division Lemma in succession until you obtain a remainder zero. HCF is the remainder in the last but one step.