LCM the 2, 3, and also 5 is the the smallest number among all usual multiples the 2, 3, and also 5. The first few multiples of 2, 3, and also 5 are (2, 4, 6, 8, 10 . . .), (3, 6, 9, 12, 15 . . .), and (5, 10, 15, 20, 25 . . .) respectively. There room 3 frequently used methods to uncover LCM that 2, 3, 5 - through listing multiples, by prime factorization, and also by division method.

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

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

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

Let's look at the different methods because that finding the LCM that 2, 3, and 5.

By element Factorization MethodBy division MethodBy Listing Multiples

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

Prime administrate of 2, 3, and also 5 is (2) = 21, (3) = 31, and (5) = 51 respectively. LCM the 2, 3, and 5 have the right to be acquired by multiply prime components raised to your respective greatest power, i.e. 21 × 31 × 51 = 30.Hence, the LCM that 2, 3, and also 5 by prime factorization is 30.

### LCM the 2, 3, and 5 by division Method To calculate the LCM of 2, 3, and also 5 by the division method, we will divide the numbers(2, 3, 5) by their prime determinants (preferably common). The product of this divisors offers the LCM that 2, 3, and 5.

Step 2: If any type of of the given numbers (2, 3, 5) is a many of 2, divide it by 2 and write the quotient listed below it. Bring down any type of number the is no divisible through the prime number.Step 3: continue the actions until only 1s room left in the critical row.

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

### LCM that 2, 3, and 5 by Listing Multiples To calculation the LCM of 2, 3, 5 through listing the end the usual multiples, we deserve to follow the given listed below steps:

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

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

☛ also Check:

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

Solution:

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

2 = 213 = 315 = 51

∴ GCD the (2, 3), (3, 5), (2, 5) and also (2, 3, 5) = 1, 1, 1 and also 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: find the smallest number the is divisible through 2, 3, 5 exactly.

Solution:

The smallest number that is divisible through 2, 3, and 5 exactly is your LCM.⇒ Multiples the 2, 3, and 5:

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

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

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go come slidego to slidego come slide ## FAQs on LCM the 2, 3, and also 5

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

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

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

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

### How to uncover the LCM the 2, 3, and 5 by element Factorization?

To uncover the LCM the 2, 3, and also 5 utilizing prime factorization, us will discover the element factors, (2 = 21), (3 = 31), and (5 = 51). LCM the 2, 3, and also 5 is the product that prime components raised to their respective highest exponent among the numbers 2, 3, and also 5.⇒ LCM of 2, 3, 5 = 21 × 31 × 51 = 30.

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

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