LCM that 2, 3, and also 5 is the the smallest number amongst all typical multiples the 2, 3, and 5. The first couple of multiples of 2, 3, and also 5 room (2, 4, 6, 8, 10 . . .), (3, 6, 9, 12, 15 . . .), and also (5, 10, 15, 20, 25 . . .) respectively. There are 3 commonly used methods to find LCM of 2, 3, 5 - by listing multiples, by prime factorization, and also by division method.

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 1 LCM the 2, 3, and also 5 2 List of Methods 3 Solved Examples 4 FAQs

Answer: LCM that 2, 3, and also 5 is 30. Explanation:

The LCM of 3 non-zero integers, a(2), b(3), and also c(5), is the smallest optimistic integer m(30) the is divisible by a(2), b(3), and also c(5) without any kind of remainder.

Let's look in ~ the various methods for finding the LCM the 2, 3, and 5.

By element Factorization MethodBy department MethodBy Listing Multiples

### LCM the 2, 3, and also 5 by prime Factorization

Prime administer of 2, 3, and also 5 is (2) = 21, (3) = 31, and (5) = 51 respectively. LCM that 2, 3, and 5 have the right to be derived by multiply prime factors raised to your respective highest possible power, i.e. 21 × 31 × 51 = 30.Hence, the LCM that 2, 3, and 5 by element factorization is 30.

### LCM that 2, 3, and also 5 by department Method To calculate the LCM the 2, 3, and also 5 by the department method, we will divide the numbers(2, 3, 5) by your prime factors (preferably common). The product of this divisors provides the LCM the 2, 3, and also 5.

Step 2: If any of the provided numbers (2, 3, 5) is a multiple of 2, divide it by 2 and also write the quotient below it. Lug down any number the is no divisible through the element number.Step 3: proceed the procedures until only 1s are left in the last row.

The LCM that 2, 3, and also 5 is the product of all prime number on the left, i.e. LCM(2, 3, 5) by department method = 2 × 3 × 5 = 30.

### LCM the 2, 3, and 5 by Listing Multiples To calculation the LCM the 2, 3, 5 by listing out the usual multiples, we have the right to follow the given listed below steps:

Step 1: list a few multiples of 2 (2, 4, 6, 8, 10 . . .), 3 (3, 6, 9, 12, 15 . . .), and 5 (5, 10, 15, 20, 25 . . .).Step 2: The typical multiples native the multiples that 2, 3, and also 5 are 30, 60, . . .Step 3: The smallest common multiple the 2, 3, and 5 is 30.

∴ The least typical multiple the 2, 3, and also 5 = 30.

Example 2: Verify the relationship between the GCD and LCM that 2, 3, and 5.

Solution:

The relation in between GCD and LCM of 2, 3, and also 5 is provided as,LCM(2, 3, 5) = <(2 × 3 × 5) × GCD(2, 3, 5)>/⇒ element factorization of 2, 3 and 5:

2 = 213 = 315 = 51

∴ GCD of (2, 3), (3, 5), (2, 5) and (2, 3, 5) = 1, 1, 1 and 1 respectively.Now, LHS = LCM(2, 3, 5) = 30.And, RHS = <(2 × 3 × 5) × GCD(2, 3, 5)>/ = <(30) × 1>/<1 × 1 × 1> = 30LHS = RHS = 30.Hence verified.

Example 3: discover the smallest number that is divisible through 2, 3, 5 exactly.

Solution:

The smallest number that is divisible by 2, 3, and 5 specifically is their LCM.⇒ Multiples the 2, 3, and 5:

Multiples the 2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, . . . .Multiples the 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, . . . .Multiples the 5 = 5, 10, 15, 20, 25, 30, 35, . . . .

Therefore, the LCM that 2, 3, and also 5 is 30.

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### What is the LCM the 2, 3, and also 5?

The LCM of 2, 3, and also 5 is 30. To uncover the LCM the 2, 3, and 5, we require to find the multiples of 2, 3, and also 5 (multiples that 2 = 2, 4, 6, 8 . . . . 30 . . . . ; multiples of 3 = 3, 6, 9, 12 . . . . 30 . . . . ; multiples that 5 = 5, 10, 15, 20, 30 . . . .) and choose the the smallest multiple that is precisely divisible by 2, 3, and also 5, i.e., 30.

### Which the the following is the LCM the 2, 3, and also 5? 3, 30, 81, 24

The worth of LCM the 2, 3, 5 is the smallest common multiple of 2, 3, and 5. The number solve the given problem is 30.

### How to uncover the LCM that 2, 3, and also 5 by prime Factorization?

To find the LCM of 2, 3, and also 5 making use of prime factorization, us will discover the prime factors, (2 = 21), (3 = 31), and also (5 = 51). LCM of 2, 3, and also 5 is the product of prime determinants raised to your respective highest exponent among the number 2, 3, and 5.⇒ LCM of 2, 3, 5 = 21 × 31 × 51 = 30.

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### What is the the very least Perfect Square Divisible by 2, 3, and 5?

The the very least number divisible through 2, 3, and also 5 = LCM(2, 3, 5)LCM of 2, 3, and 5 = 2 × 3 × 5 ⇒ least perfect square divisible by each 2, 3, and also 5 = LCM(2, 3, 5) × 2 × 3 × 5 = 900 Therefore, 900 is the forced number.