Finding the Lowest Common Multiple of Whole Numbers for JSS 1

Lowest Common Multiples (LCM)

Performance Objectives

By the end of the lesson, students should be able to:

1. identify multiples and common multiples of two or more whole numbers;
2. find the LCM of two or more whole numbers;
3. determine common multiples of whole numbers.

Multiples of Whole Numbers

A multiple of a whole number is the product obtained when that number is multiplied by any other whole number.

To find the multiples of any whole number, the number is multiplied by $1, 2, 3, 4, 5,$ and so on. This process is similar to writing out the multiplication table of that number. Multiples of any number are infinite, meaning they continue without end.

For example, the multiples of $3$ are obtained as follows:

$$3 \times 1 = 3$$

$$3 \times 2 = 6$$

$$3 \times 3 = 9$$

$$3 \times 4 = 12$$

$$3 \times 5 = 15$$

The list of multiples of $3$ is written as: $3, 6, 9, 12, 15, 18, 21, \dots$

The list of multiples of $4$ is written as: $4, 8, 12, 16, 20, 24, 28, \dots$

Common Multiples of Whole Numbers

Common multiples of two or more whole numbers are numbers that are multiples of each of the given numbers.

When the multiples of two or more different numbers are listed, some numbers will appear in all the lists. These shared numbers are the common multiples.

To find the common multiples of $4$ and $6$, the multiples of each number are listed first:

The multiples of $4$ are:

$$4, 8, \mathbf{12}, 16, 20, \mathbf{24}, 28, 32, \mathbf{36}, 40, \dots$$

The multiples of $6$ are:

$$6, \mathbf{12}, 18, \mathbf{24}, 30, \mathbf{36}, 42, \dots$$

The numbers that appear in both lists are $12, 24,$ and $36$. Therefore, the common multiples of $4$ and $6$ are $12, 24, 36, \dots$

The Lowest Common Multiple (LCM)

The Lowest Common Multiple (LCM) of two or more whole numbers is the smallest non-zero number that is a multiple of all the given numbers.

From the common multiples of $4$ and $6$ listed above ($12, 24, 36, \dots$), the smallest number is $12$. Therefore, the Lowest Common Multiple (LCM) of $4$ and $6$ is $12$.

Methods of Finding the LCM

There are three main methods used to find the Lowest Common Multiple of whole numbers:

1. The Multiple Method (Listing Multiples)
2. The Factor Method (Continuous Division)
3. The Index Method (Prime Factorization)

Method 1: The Multiple Method (Listing Multiples)

This method involves listing the multiples of each number, identifying the common multiples, and selecting the smallest one. This method is most suitable for smaller numbers.

The procedure involves the following steps:

1. List the first few multiples of each given number.
2. Identify and circle the common multiples.
3. Select the smallest of these common multiples as the LCM.

Worked Example 1

Find the LCM of $6$ and $8$ using the multiple method.

Solution:

First, list the multiples of each number:

The multiples of $6$ are:

$$6, 12, 18, \mathbf{24}, 30, 36, 42, \mathbf{48}, 54, \dots$$

The multiples of $8$ are:

$$8, 16, \mathbf{24}, 32, 40, \mathbf{48}, 56, \dots$$

The common multiples of $6$ and $8$ are $24, 48, \dots$

The smallest of these common multiples is $24$.

Therefore, the LCM of $6$ and $8$ is $24$.

Worked Example 2

Find the LCM of $3, 5,$ and $6$ using the multiple method.

Solution:

List the multiples of each of the three numbers:

The multiples of $3$ are:

$$3, 6, 9, 12, 15, 18, 21, 24, 27, \mathbf{30}, 33, 36, \dots$$

The multiples of $5$ are:

$$5, 10, 15, 20, 25, \mathbf{30}, 35, 40, \dots$$

The multiples of $6$ are:

$$6, 12, 18, 24, \mathbf{30}, 36, 42, \dots$$

The smallest multiple that appears in all three lists is $30$.

Therefore, the LCM of $3, 5,$ and $6$ is $30$.

Method 2: The Factor Method (Continuous Division)

This method uses a division table to divide the given numbers by prime factors simultaneously. It is highly efficient for larger numbers or when finding the LCM of three or more numbers.

The procedure involves the following steps:

1. Arrange the numbers in a row inside a division table.
2. Divide the numbers by the smallest prime number (starting with $2, 3, 5, 7, \dots$) that can divide at least one of the numbers.
3. Write the quotient directly below each number. If a number is not exactly divisible by the prime number, write the number down unchanged in the next row.
4. Repeat the division process using prime numbers until the quotient for all numbers in the bottom row is $1$.
5. Multiply all the prime factors in the divisor column together to obtain the LCM.

Worked Example 1

Find the LCM of $12$ and $18$ using the factor method.

Solution:

Set up the division table using prime numbers:

Multiply the prime divisors in the left column:

$$\text{LCM} = 2 \times 2 \times 3 \times 3$$

$$\text{LCM} = 4 \times 9 = 36$$

Therefore, the LCM of $12$ and $18$ is $36$.

Worked Example 2

Find the LCM of $15, 20,$ and $30$ using the factor method.

Solution:

Set up the division table for the three numbers:

Multiply the prime divisors in the left column:

$$\text{LCM} = 2 \times 2 \times 3 \times 5$$

$$\text{LCM} = 4 \times 3 \times 5$$

$$\text{LCM} = 12 \times 5 = 60$$

Therefore, the LCM of $15, 20,$ and $30$ is $60$.

Method 3: The Index Method (Prime Factorization)

This method involves expressing each number as a product of its prime factors written in index form (using powers or exponents). The LCM is then found by multiplying the highest power of each prime factor that appears in the expressions.

The procedure involves the following steps:

1. Express each number as a product of prime factors.
2. Write the prime factors in index form.
3. Identify all the unique prime bases present across all the numbers.
4. For each unique prime base, choose the term with the highest power (highest index).
5. Multiply these highest powers together to get the LCM.

Worked Example 1

Find the LCM of $12$ and $18$ using the index method.

Solution:

First, express each number as a product of prime factors:

$$12 = 2 \times 2 \times 3$$

$$18 = 2 \times 3 \times 3$$

Next, write them in index form:

$$12 = 2^2 \times 3^1$$

$$18 = 2^1 \times 3^2$$

Identify the unique prime bases, which are $2$ and $3$.

Select the highest power for each prime base:

The highest power of $2$ is $2^2$.

The highest power of $3$ is $3^2$.

Multiply these highest powers together:

$$\text{LCM} = 2^2 \times 3^2$$

$$\text{LCM} = 4 \times 9 = 36$$

Therefore, the LCM of $12$ and $18$ is $36$.

Worked Example 2

Find the LCM of $24, 36,$ and $40$ using the index method.

Solution:

Express each number as a product of prime factors in index form:

$$24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1$$

$$36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$$

$$40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5^1$$

Identify the unique prime bases, which are $2, 3,$ and $5$.

Select the highest power for each prime base:

The highest power of $2$ is $2^3$.

The highest power of $3$ is $3^2$.

The highest power of $5$ is $5^1$.

Multiply these highest powers together:

$$\text{LCM} = 2^3 \times 3^2 \times 5^1$$

$$\text{LCM} = 8 \times 9 \times 5$$

$$\text{LCM} = 72 \times 5 = 360$$

Therefore, the LCM of $24, 36,$ and $40$ is $360$.

Practical Applications of LCM

The concept of Lowest Common Multiple is used to solve real-life problems where events occur repeatedly at different regular intervals, or when items must be grouped equally without any remainder.

Worked Example 1

Two bells in a school ring at intervals of $15$ minutes and $20$ minutes respectively. If they ring together at 8:00 AM, at what time will they ring together again?

Solution:

To find when the bells will ring together again, find the LCM of the two time intervals, $15$ and $20$.

Using the index method:

$$15 = 3^1 \times 5^1$$

$$20 = 2^2 \times 5^1$$

The unique prime bases are $2, 3,$ and $5$.

Select the highest power of each base:

The highest power of $2$ is $2^2$.

The highest power of $3$ is $3^1$.

The highest power of $5$ is $5^1$.

Multiply these values:

$$\text{LCM} = 2^2 \times 3^1 \times 5^1$$

$$\text{LCM} = 4 \times 3 \times 5 = 60\text{ minutes}$$

Since $60\text{ minutes} = 1\text{ hour}$, the bells will ring together again after exactly $1$ hour.

Add $1$ hour to the starting time of 8:00 AM:

$$\text{New Time} = \text{8:00 AM} + 1\text{ hour} = \text{9:00 AM}$$

Therefore, the bells will ring together again at 9:00 AM.

Worked Example 2

A trader in a local market wants to pack oranges into bags. If she packs them in groups of $12$ or $18$, she has no oranges left over. What is the smallest number of oranges she could have?

Solution:

The smallest number of oranges must be a multiple of both $12$ and $18$. Therefore, find the LCM of $12$ and $18$.

Using the factor method:

$$\text{LCM} = 2 \times 2 \times 3 \times 3 = 36$$

Therefore, the smallest number of oranges the trader could have is $36$.