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HCF of 2, 3 and 5

HCF of 2, 3 and 5 is the largest possible number that divides 2, 3 and 5 exactly without any remainder. The factors of 2, 3 and 5 are (1, 2), (1, 3) and (1, 5) respectively. There are 3 commonly used methods to find the HCF of 2, 3 and 5 - prime factorization, Euclidean algorithm, and long division.

1.	HCF of 2, 3 and 5
2.	List of Methods
3.	Solved Examples
4.	FAQs

What is HCF of 2, 3 and 5?

Answer: HCF of 2, 3 and 5 is 1.

Explanation:

The HCF of three non-zero integers, x(2), y(3) and z(5), is the highest positive integer m(1) that divides x(2), y(3) and z(5) without any remainder.

Methods to Find HCF of 2, 3 and 5

The methods to find the HCF of 2, 3 and 5 are explained below.

Prime Factorization Method
Using Euclid's Algorithm
Listing Common Factors

HCF of 2, 3 and 5 by Prime Factorization

Prime factorization of 2, 3 and 5 is (2), (3) and (5) respectively. As visible, there are no common prime factors between 2, 3 and 5, i.e. they are co-prime. Hence, the HCF of 2, 3 and 5 will be 1.

HCF of 2, 3 and 5 by Euclidean Algorithm

As per the Euclidean Algorithm, HCF(X, Y) = HCF(Y, X mod Y)
where X > Y and mod is the modulo operator.

HCF(2, 3, 5) = HCF(HCF(2, 3), 5)

HCF(3, 2) = HCF(2, 3 mod 2) = HCF(2, 1)
HCF(2, 1) = HCF(1, 2 mod 1) = HCF(1, 0)
HCF(1, 0) = 1 (∵ HCF(X, 0) = |X|, where X ≠ 0)

Steps for HCF(1, 5)

HCF(5, 1) = HCF(1, 5 mod 1) = HCF(1, 0)
HCF(1, 0) = 1 (∵ HCF(X, 0) = |X|, where X ≠ 0)

Therefore, the value of HCF of 2, 3 and 5 is 1.

HCF of 2, 3 and 5 by Listing Common Factors

Factors of 2: 1, 2
Factors of 3: 1, 3
Factors of 5: 1, 5

Since, 1 is the only common factor between 2, 3 and 5. The Highest Common Factor of 2, 3 and 5 is 1.

☛ Also Check:

HCF of 2, 3 and 5 Examples

Example 1: Verify the relation between the LCM and HCF of 2, 3 and 5.

Solution:

The relation between the LCM and HCF of 2, 3 and 5 is given as, HCF(2, 3, 5) = [(2 × 3 × 5) × LCM(2, 3, 5)]/[LCM(2, 3) × LCM (3, 5) × LCM(2, 5)]
⇒ Prime factorization of 2, 3 and 5:
- 2 = 2
- 3 = 3
- 5 = 5
∴ LCM of (2, 3), (3, 5), (2, 5), and (2, 3, 5) is 6, 15, 10, and 30 respectively.
Now, LHS = HCF(2, 3, 5) = 1.
And, RHS = [(2 × 3 × 5) × LCM(2, 3, 5)]/[LCM(2, 3) × LCM (3, 5) × LCM(2, 5)] = [(30) × 30]/[6 × 15 × 10]
LHS = RHS = 1.
Hence verified.
Example 2: Find the highest number that divides 2, 3, and 5 completely.

Solution:

The highest number that divides 2, 3, and 5 exactly is their highest common factor.
- Factors of 2 = 1, 2
- Factors of 3 = 1, 3
- Factors of 5 = 1, 5
The HCF of 2, 3, and 5 is 1.
∴ The highest number that divides 2, 3, and 5 is 1.
Example 3: Calculate the HCF of 2, 3, and 5 using LCM of the given numbers.

Solution:

Prime factorization of 2, 3 and 5 is given as,
- 2 = 2
- 3 = 3
- 5 = 5
LCM(2, 3) = 6, LCM(3, 5) = 15, LCM(5, 2) = 10, LCM(2, 3, 5) = 30
⇒ HCF(2, 3, 5) = [(2 × 3 × 5) × LCM(2, 3, 5)]/[LCM(2, 3) × LCM (3, 5) × LCM(5, 2)]
⇒ HCF(2, 3, 5) = (30 × 30)/(6 × 15 × 10)
⇒ HCF(2, 3, 5) = 1.
Therefore, the HCF of 2, 3 and 5 is 1.

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FAQs on HCF of 2, 3 and 5

What is the HCF of 2, 3 and 5?

The HCF of 2, 3 and 5 is 1. To calculate the highest common factor of 2, 3 and 5, we need to factor each number (factors of 2 = 1, 2; factors of 3 = 1, 3; factors of 5 = 1, 5) and choose the highest factor that exactly divides 2, 3 and 5, i.e., 1.

Which of the following is HCF of 2, 3 and 5? 1, 49, 37, 18, 27, 23, 28

HCF of 2, 3, 5 will be the number that divides 2, 3, and 5 without leaving any remainder. The only number that satisfies the given condition is 1.

What are the Methods to Find HCF of 2, 3 and 5?

There are three commonly used methods to find the HCF of 2, 3 and 5.

By Euclidean Algorithm
By Prime Factorization
By Long Division

What is the Relation Between LCM and HCF of 2, 3 and 5?

The following equation can be used to express the relation between LCM and HCF of 2, 3 and 5, i.e. HCF(2, 3, 5) = [(2 × 3 × 5) × LCM(2, 3, 5)]/[LCM(2, 3) × LCM (3, 5) × LCM(2, 5)].
☛ Highest Common Factor Calculator

How to Find the HCF of 2, 3 and 5 by Prime Factorization?

To find the HCF of 2, 3 and 5, we will find the prime factorization of given numbers, i.e. 2 = 2; 3 = 3; 5 = 5.
⇒ There is no common prime factor for 2, 3 and 5. Hence, HCF(2, 3, 5) = 1.
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