Highest Common Factor of 112, 483 using Euclid's algorithm
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 112, 483 i.e. 7 the largest integer that leaves a remainder zero for all numbers.
HCF of 112, 483 is 7 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 112, 483 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 112, 483 is 7.
HCF(112, 483) = 7
HCF of 112, 483 using Euclid's algorithm
Highest common factor or Highest common divisor (hcd) can be calculated by Euclid's algotithm.
Highest common factor (HCF) of 112, 483 is 7.
Highest Common Factor of 112,483 is 7
Step 1: Since 483 > 112, we apply the division lemma to 483 and 112, to get
483 = 112 x 4 + 35
Step 2: Since the reminder 112 ≠ 0, we apply division lemma to 35 and 112, to get
112 = 35 x 3 + 7
Step 3: We consider the new divisor 35 and the new remainder 7, and apply the division lemma to get
35 = 7 x 5 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 7, the HCF of 112 and 483 is 7
Notice that 7 = HCF(35,7) = HCF(112,35) = HCF(483,112) .
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Frequently Asked Questions on HCF of 112, 483 using Euclid's Algorithm
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 112, 483?
Answer: HCF of 112, 483 is 7 the largest number that divides all the numbers leaving a remainder zero.
3. How to find HCF of 112, 483 using Euclid's Algorithm?
Answer: For arbitrary numbers 112, 483 apply Euclid’s Division Lemma in succession until you obtain a remainder zero. HCF is the remainder in the last but one step.