Prime Factorization and the Least Common Multiple

Find the Prime Factorization of a Composite Number

In the previous section, we found the factors of a number. Prime numbers have only two factors, the number $1$ and the prime number itself. Composite numbers have more than two factors, and every composite number can be written as a unique product of primes. This is called the prime factorization of a number. When we write the prime factorization of a number, we are rewriting the number as a product of primes. Finding the prime factorization of a composite number will help you later in this course.

You may want to refer to the following list of prime numbers less than $50$ as you work through this section.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Prime Factorization Using the Factor Tree Method

One way to find the prime factorization of a number is to make a factor tree. We start by writing the number, and then writing it as the product of two factors. We write the factors below the number and connect them to the number with a small line segment—a “branch” of the factor tree.

If a factor is prime, we circle it (like a bud on a tree), and do not factor that “branch” any further. If a factor is not prime, we repeat this process, writing it as the product of two factors and adding new branches to the tree.

We continue until all the branches end with a prime. When the factor tree is complete, the circled primes give us the prime factorization.

For example, let’s find the prime factorization of $36 .$ We can start with any factor pair such as $3$ and $12 .$ We write $3$ and $12$ below $36$ with branches connecting them.

The figure shows a factor tree with the number 36 at the top. Two branches are splitting out from under 36. The right branch has a number 3 at the end with a circle around it. The left branch has the number 12 at the end.

The factor $3$ is prime, so we circle it. The factor $12$ is composite, so we need to find its factors. Let’s use $3$ and $4 .$ We write these factors on the tree under the $12 .$

The factor $3$ is prime, so we circle it. The factor $4$ is composite, and it factors into $2 \cdot 2 .$ We write these factors under the $4 .$ Since $2$ is prime, we circle both $2 s .$

The prime factorization is the product of the circled primes. We generally write the prime factorization in order from least to greatest.

2 \cdot 2 \cdot 3 \cdot 3

In cases like this, where some of the prime factors are repeated, we can write prime factorization in exponential form.

\begin{matrix} 2 \cdot 2 \cdot 3 \cdot 3 \\ 2^{2} \cdot 3^{2} \end{matrix}

Note that we could have started our factor tree with any factor pair of $36 .$ We chose $12$ and $3,$ but the same result would have been the same if we had started with $2$ and $18, 4$ and $9, or 6 and 6 .$

Find the prime factorization of $48$ using the factor tree method.

Solution

We can start our tree using any factor pair of 48. Let's use 2 and 24. We circle the 2 because it is prime and so that branch is complete.
Now we will factor 24. Let's use 4 and 6.
Neither factor is prime, so we do not circle either. We factor the 4, using 2 and 2. We factor 6, using 2 and 3. We circle the 2s and the 3 since they are prime. Now all of the branches end in a prime.
Write the product of the circled numbers.	$2 \cdot 2 \cdot 2 \cdot 2 \cdot 3$
Write in exponential form.	$2^{4} \cdot 3$

Check this on your own by multiplying all the factors together. The result should be $48 .$

Find the prime factorization of 84 using the factor tree method.

Solution

We start with the factor pair 4 and 21.

Neither factor is prime so we factor them further.

A factor tree shows the number 84 branching into its factors 4 and 21.

Now the factors are all prime, so we circle them.

A prime factorization tree illustrating how the number 84 breaks down into its prime factors: 2, 2, 3, and 7.

Then we write 84 as the product of all circled primes.

2 \cdot 2 \cdot 3 \cdot 7

2^{2} \cdot 3 \cdot 7

Draw a factor tree of $84 .$

Prime Factorization Using the Ladder Method

The ladder method is another way to find the prime factors of a composite number. It leads to the same result as the factor tree method. Some people prefer the ladder method to the factor tree method, and vice versa.

To begin building the “ladder,” divide the given number by its smallest prime factor. For example, to start the ladder for $36,$ we divide $36$ by $2,$ the smallest prime factor of $36 .$

The image shows the division of 2 into 36 to get the quotient 18. This division is represented using a division bracket with 2 on the outside left of the bracket, 36 inside the bracket and 18 above the 36, outside the bracket.

To add a “step” to the ladder, we continue dividing by the same prime until it no longer divides evenly.

Then we divide by the next prime; so we divide $9$ by $3 .$

We continue dividing up the ladder in this way until the quotient is prime. Since the quotient, $3,$ is prime, we stop here.

Do you see why the ladder method is sometimes called stacked division?

The prime factorization is the product of all the primes on the sides and top of the ladder.

\begin{matrix} 2 \cdot 2 \cdot 3 \cdot 3 \\ 2^{2} \cdot 3^{2} \end{matrix}

Notice that the result is the same as we obtained with the factor tree method.

Find the prime factorization of $120$ using the ladder method.

Solution

Divide the number by the smallest prime, which is 2.
Continue dividing by 2 until it no longer divides evenly.
Divide by the next prime, 3.
The quotient, 5, is prime, so the ladder is complete. Write the prime factorization of 120.	$2 \cdot 2 \cdot 2 \cdot 3 \cdot 5$ $2^{3} \cdot 3 \cdot 5$

Check this yourself by multiplying the factors. The result should be $120 .$

Find the prime factorization of $48$ using the ladder method.

Solution

Divide the number by the smallest prime, 2.
Continue dividing by 2 until it no longer divides evenly.
The quotient, 3, is prime, so the ladder is complete. Write the prime factorization of 48.	$2 \cdot 2 \cdot 2 \cdot 2 \cdot 3$ $2^{4} \cdot 3$

Find the Least Common Multiple (LCM) of Two Numbers

One of the reasons we look at multiples and primes is to use these techniques to find the least common multiple of two numbers. This will be useful when we add and subtract fractions with different denominators.

Listing Multiples Method

A common multiple of two numbers is a number that is a multiple of both numbers. Suppose we want to find common multiples of $10$ and $25 .$ We can list the first several multiples of each number. Then we look for multiples that are common to both lists—these are the common multiples.

\begin{matrix} 10 : 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, … \\ 25 : 25, 50, 75, 100, 125, … \end{matrix}

We see that $50$ and $100$ appear in both lists. They are common multiples of $10$ and $25 .$ We would find more common multiples if we continued the list of multiples for each.

The smallest number that is a multiple of two numbers is called the least common multiple (LCM). So the least LCM of $10$ and $25$ is $50 .$

Find the LCM of $15$ and $20$ by listing multiples.

Solution

List the first several multiples of $15$ and of $20 .$ Identify the first common multiple.

$\begin{array}{l} 15: 15, 30, 45, 60, 75, 90, 105, 120 \\ 20: 20, 40, 60, 80, 100, 120, 140, 160 \end{array}$

The smallest number to appear on both lists is $60,$ so $60$ is the least common multiple of $15$ and $20 .$

Notice that $120$ is on both lists, too. It is a common multiple, but it is not the least common multiple.

Prime Factors Method

Another way to find the least common multiple of two numbers is to use their prime factors. We’ll use this method to find the LCM of $12$ and $18 .$

We start by finding the prime factorization of each number.

12 = 2 \cdot 2 \cdot 3 18 = 2 \cdot 3 \cdot 3

Then we write each number as a product of primes, matching primes vertically when possible.

\begin{array}{l} 12 = 2 \cdot 2 \cdot 3 \\ 18 = 2 \cdot 3 \cdot 3 \end{array}

Now we bring down the primes in each column. The LCM is the product of these factors.

The image shows the prime factorization of 12 written as the equation 12 equals 2 times 2 times 3. Below this equation is another showing the prime factorization of 18 written as the equation 18 equals 2 times 3 times 3. The two equations line up vertically at the equal symbol. The first 2 in the prime factorization of 12 aligns with the 2 in the prime factorization of 18. Under the second 2 in the prime factorization of 12 is a gap in the prime factorization of 18. Under the 3 in the prime factorization of 12 is the first 3 in the prime factorization of 18. The second 3 in the prime factorization has no factors above it from the prime factorization of 12. A horizontal line is drawn under the prime factorization of 18. Below this line is the equation LCM equal to 2 times 2 times 3 times 3. Arrows are drawn down vertically from the prime factorization of 12 through the prime factorization of 18 ending at the LCM equation. The first arrow starts at the first 2 in the prime factorization of 12 and continues down through the 2 in the prime factorization of 18. Ending with the first 2 in the LCM. The second arrow starts at the next 2 in the prime factorization of 12 and continues down through the gap in the prime factorization of 18. Ending with the second 2 in the LCM. The third arrow starts at the 3 in the prime factorization of 12 and continues down through the first 3 in the prime factorization of 18. Ending with the first 3 in the LCM. The last arrow starts at the second 3 in the prime factorization of 18 and points down to the second 3 in the LCM.

Notice that the prime factors of $12$ and the prime factors of $18$ are included in the LCM. By matching up the common primes, each common prime factor is used only once. This ensures that $36$ is the least common multiple.

Find the LCM of $15$ and $18$ using the prime factors method.

Solution

Write each number as a product of primes.
Write each number as a product of primes, matching primes vertically when possible.
Bring down the primes in each column.
Multiply the factors to get the LCM.	$LCM = 2 \cdot 3 \cdot 3 \cdot 5$ The LCM of 15 and 18 is 90.

Find the LCM of $50$ and $100$ using the prime factors method.

Solution

Write the prime factorization of each number.
Write each number as a product of primes, matching primes vertically when possible.
Bring down the primes in each column.
Multiply the factors to get the LCM.	$LCM = 2 \cdot 2 \cdot 5 \cdot 5$ The LCM of 50 and 100 is 100.

Key Concepts

Find the prime factorization of a composite number using the tree method.
1. Find any factor pair of the given number, and use these numbers to create two branches.
2. If a factor is prime, that branch is complete. Circle the prime.
3. If a factor is not prime, write it as the product of a factor pair and continue the process.
4. Write the composite number as the product of all the circled primes.
Find the prime factorization of a composite number using the ladder method.
1. Divide the number by the smallest prime.
2. Continue dividing by that prime until it no longer divides evenly.
3. Divide by the next prime until it no longer divides evenly.
4. Continue until the quotient is a prime.
5. Write the composite number as the product of all the primes on the sides and top of the ladder.
Find the LCM by listing multiples.
1. List the first several multiples of each number.
2. Look for multiples common to both lists. If there are no common multiples in the lists, write out additional multiples for each number.
3. Look for the smallest number that is common to both lists.
4. This number is the LCM.
Find the LCM using the prime factors method.
1. Find the prime factorization of each number.
2. Write each number as a product of primes, matching primes vertically when possible.
3. Bring down the primes in each column.
4. Multiply the factors to get the LCM.

Section Exercises

Practice Makes Perfect

Find the Prime Factorization of a Composite Number

In the following exercises, find the prime factorization of each number using the factor tree method.

$86$

Solution

2 ⋅ 43

$78$

$132$

Solution

2 ⋅ 2 ⋅ 3 ⋅ 11

$455$

$693$

Solution

3 ⋅ 3 ⋅ 7 ⋅ 11

$420$

$115$

Solution

5 ⋅ 23

$225$

$2475$

Solution

3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 11

1560

In the following exercises, find the prime factorization of each number using the ladder method.

$56$

Solution

2 ⋅ 2 ⋅ 2 ⋅ 7

$72$

$168$

Solution

2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 7

$252$

$391$

Solution

17 ⋅ 23

$400$

$432$

Solution

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3

$627$

$2160$

Solution

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 5

$2520$

In the following exercises, find the prime factorization of each number using any method.

$150$

Solution

2 ⋅ 3 ⋅ 5 ⋅ 5

$180$

$525$

Solution

3 ⋅ 5 ⋅ 5 ⋅ 7

$444$

$36$

Solution

2 ⋅ 2 ⋅ 3 ⋅ 3

$50$

$350$

Solution

2 ⋅ 5 ⋅ 5 ⋅ 7

$144$

Find the Least Common Multiple (LCM) of Two Numbers

In the following exercises, find the least common multiple (LCM) by listing multiples.

$8, 12$

Solution

24

$4, 3$

$6, 15$

Solution

30

$12, 16$

$30, 40$

Solution

120

$20, 30$

$60, 75$

Solution

300

$44, 55$

In the following exercises, find the least common multiple (LCM) by using the prime factors method.

$8, 12$

Solution

24

$12, 16$

$24, 30$

Solution

120

$28, 40$

$70, 84$

Solution

420

$84, 90$

In the following exercises, find the least common multiple (LCM) using any method.

$6, 21$

Solution

42

$9, 15$

$24, 30$

Solution

120

$32, 40$

Everyday Math

Grocery shopping Hot dogs are sold in packages of ten, but hot dog buns come in packs of eight. What is the smallest number of hot dogs and buns that can be purchased if you want to have the same number of hot dogs and buns? (Hint: it is the LCM!)

Solution

40

Grocery shopping Paper plates are sold in packages of $12$ and party cups come in packs of $8 .$ What is the smallest number of plates and cups you can purchase if you want to have the same number of each? (Hint: it is the LCM!)

Writing Exercises

Do you prefer to find the prime factorization of a composite number by using the factor tree method or the ladder method? Why?

Do you prefer to find the LCM by listing multiples or by using the prime factors method? Why?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

A self-assessment table for math skills: prime factorization and least common multiple (LCM). Users can rate their understanding as Confidently, With some help, or No-I don't get it!

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Use the Language of Algebra

Use Variables and Algebraic Symbols

In the following exercises, translate from algebra to English.

$3 \cdot 8$

Solution

3 times 8, the product of three and eight.

$12 - x$

$24 \div 6$

Solution

24 divided by 6, the quotient of twenty-four and six.

$9 + 2 a$

$50 \geq 47$

Solution

50 is greater than or equal to 47

$3 y < 15$

$n + 4 = 13$

Solution

The sum of n and 4 is equal to 13

$32 - k = 7$

Identify Expressions and Equations

In the following exercises, determine if each is an expression or equation.

$5 + u = 84$

Solution

equation

$36 - 6 s$

$4 y - 11$

Solution

expression

$10 x = 120$

Simplify Expressions with Exponents

In the following exercises, write in exponential form.

$2 \cdot 2 \cdot 2$

Solution

2³

$a \cdot a \cdot a \cdot a \cdot a$

$x \cdot x \cdot x \cdot x \cdot x \cdot x$

Solution

x⁶

$10 \cdot 10 \cdot 10$

In the following exercises, write in expanded form.

$8^{4}$

Solution

8 ⋅ 8 ⋅ 8 ⋅ 8

$3^{6}$

$y^{5}$

Solution

y ⋅ y ⋅ y ⋅ y ⋅ y

$n^{4}$

In the following exercises, simplify each expression.

$3^{4}$

Solution

81

$10^{6}$

$2^{7}$

Solution

128

$4^{3}$

Simplify Expressions Using the Order of Operations

In the following exercises, simplify.

$10 + 2 \cdot 5$

Solution

20

$(10 + 2) \cdot 5$

$(30 + 6) \div 2$

Solution

18

$30 + 6 \div 2$

$7^{2} + 5^{2}$

Solution

74

${(7 + 5)}^{2}$

$4 + 3 (10 - 1)$

Solution

31

$(4 + 3) (10 - 1)$

Evaluate, Simplify, and Translate Expressions

Evaluate an Expression

In the following exercises, evaluate the following expressions.

$9 x - 5 when x = 7$

Solution

58

$y^{3} when y = 5$

$3 a - 4 b when a = 10, b = 1$

Solution

26

$b h when b = 7, h = 8$

Identify Terms, Coefficients and Like Terms

In the following exercises, identify the terms in each expression.

$12 n^{2} + 3 n + 1$

Solution

12n²,3n, 1

$4 x^{3} + 11 x + 3$

In the following exercises, identify the coefficient of each term.

$6 y$

Solution

6

$13 x^{2}$

In the following exercises, identify the like terms.

$5 x^{2}, 3, 5 y^{2}, 3 x, x, 4$

Solution

3 and 4; 3x and x

$8, 8 r^{2}, 8 r, 3 r, r^{2}, 3 s$

Simplify Expressions by Combining Like Terms

In the following exercises, simplify the following expressions by combining like terms.

$15 a + 9 a$

Solution

24a

$12 y + 3 y + y$

$4 x + 7 x + 3 x$

Solution

14x

$6 + 5 c + 3$

$8 n + 2 + 4 n + 9$

Solution

12n + 11

$19 p + 5 + 4 p - 1 + 3 p$

$7 y^{2} + 2 y + 11 + 3 y^{2} - 8$

Solution

10y² + 2y + 3

$13 x^{2} - x + 6 + 5 x^{2} + 9 x$

Translate English Phrases to Algebraic Expressions

In the following exercises, translate the following phrases into algebraic expressions.

the difference of $x$ and $6$

Solution

x − 6

the sum of $10$ and twice $a$

the product of $3 n$ and $9$

Solution

3n ⋅ 9

the quotient of $s$ and $4$

$5$ times the sum of $y$ and $1$

Solution

5(y + 1)

$10$ less than the product of $5$ and $z$

Jack bought a sandwich and a coffee. The cost of the sandwich was $$3$ more than the cost of the coffee. Call the cost of the coffee $c .$ Write an expression for the cost of the sandwich.

Solution

c + 3

The number of poetry books on Brianna’s bookshelf is $5$ less than twice the number of novels. Call the number of novels $n .$ Write an expression for the number of poetry books.

Solve Equations Using the Subtraction and Addition Properties of Equality

Determine Whether a Number is a Solution of an Equation

In the following exercises, determine whether each number is a solution to the equation.

y + 16 = 40

ⓐ $24$
ⓑ $56$

Solution

ⓐ yes
ⓑ no

d - 6 = 21

ⓐ $15$
ⓑ $27$

4 n + 12 = 36

ⓐ $6$
ⓑ $12$

Solution

ⓐ yes
ⓑ no

20 q - 10 = 70

ⓐ $3$
ⓑ $4$

15 x - 5 = 10 x + 45

ⓐ $2$
ⓑ $10$

Solution

ⓐ no
ⓑ yes

22 p - 6 = 18 p + 86

ⓐ $4$
ⓑ $23$

Model the Subtraction Property of Equality

In the following exercises, write the equation modeled by the envelopes and counters and then solve the equation using the subtraction property of equality.

This image is divided into two parts: the first part shows an envelope and 3 blue counters and the next to it, the second part shows five counters.

Solution

x + 3 = 5; x = 2

This image is divided into two parts: the first part shows an envelope and 4 blue counters and next to it, the second part shows 9 counters.

Solve Equations using the Subtraction Property of Equality

In the following exercises, solve each equation using the subtraction property of equality.

$c + 8 = 14$

Solution

c = 6

$v + 8 = 150$

$23 = x + 12$

Solution

x = 11

$376 = n + 265$

Solve Equations using the Addition Property of Equality

In the following exercises, solve each equation using the addition property of equality.

$y - 7 = 16$

Solution

y = 23

$k - 42 = 113$

$19 = p - 15$

Solution

p = 34

$501 = u - 399$

Translate English Sentences to Algebraic Equations

In the following exercises, translate each English sentence into an algebraic equation.

The sum of $7$ and $33$ is equal to $40 .$

Solution

7 + 33 = 40

The difference of $15$ and $3$ is equal to $12 .$

The product of $4$ and $8$ is equal to $32 .$

Solution

4 ⋅ 8 = 32

The quotient of $63$ and $9$ is equal to $7 .$

Twice the difference of $n$ and $3$ gives $76 .$

Solution

2(n − 3) = 76

The sum of five times $y$ and $4$ is $89 .$

Translate to an Equation and Solve

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

Eight more than $x$ is equal to $35 .$

Solution

x + 8 = 35; x = 27

$21$ less than $a$ is $11 .$

The difference of $q$ and $18$ is $57 .$

Solution

q − 18 = 57; q = 75

The sum of $m$ and $125$ is $240 .$

Mixed Practice

In the following exercises, solve each equation.

$h - 15 = 27$

Solution

h = 42

$k - 11 = 34$

$z + 52 = 85$

Solution

z = 33

$x + 93 = 114$

$27 = q + 19$

Solution

q = 8

$38 = p + 19$

$31 = v - 25$

Solution

v = 56

$38 = u - 16$

Find Multiples and Factors

Identify Multiples of Numbers

In the following exercises, list all the multiples less than $50$ for each of the following.

$3$

Solution

3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48

$2$

$8$

Solution

8, 16, 24, 32, 40, 48

$10$

Use Common Divisibility Tests

In the following exercises, using the divisibility tests, determine whether each number is divisible by $2, by 3, by 5, by 6, and by 10 .$

$96$

Solution

2, 3, 6

$250$

$420$

Solution

2, 3, 5, 6, 10

$625$

Find All the Factors of a Number

In the following exercises, find all the factors of each number.

$30$

Solution

1, 2, 3, 5, 6, 10, 15, 30

$70$

$180$

Solution

1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180

$378$

Identify Prime and Composite Numbers

In the following exercises, identify each number as prime or composite.

$19$

Solution

prime

$51$

$121$

Solution

composite

$219$

Prime Factorization and the Least Common Multiple

Find the Prime Factorization of a Composite Number

In the following exercises, find the prime factorization of each number.

$84$

Solution

2 ⋅ 2 ⋅ 3 ⋅ 7

$165$

$350$

Solution

2 ⋅ 5 ⋅ 5 ⋅ 7

$572$

Find the Least Common Multiple of Two Numbers

In the following exercises, find the least common multiple of each pair of numbers.

$9, 15$

Solution

45

$12, 20$

$25, 35$

Solution

175

$18, 40$

Everyday Math

Describe how you have used two topics from The Language of Algebra chapter in your life outside of your math class during the past month.

Solution

Answers will vary

Chapter Practice Test

In the following exercises, translate from an algebraic equation to English phrases.

$6 \cdot 4$

$15 - x$

Solution

15 minus x, the difference of fifteen and x.

In the following exercises, identify each as an expression or equation.

$5 \cdot 8 + 10$

$x + 6 = 9$

Solution

equation

$3 \cdot 11 = 33$

ⓐ Write $n \cdot n \cdot n \cdot n \cdot n \cdot n$ in exponential form.
ⓑ Write $3^{5}$ in expanded form and then simplify.

Solution

ⓐ n⁶
ⓑ 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3 = 243

In the following exercises, simplify, using the order of operations.

$4 + 3 \cdot 5$

$(8 + 1) \cdot 4$

Solution

36

$1 + 6 (3 - 1)$

$(8 + 4) \div 3 + 1$

Solution

5

${(1 + 4)}^{2}$

$5 [2 + 7 (9 - 8)]$

Solution

45

In the following exercises, evaluate each expression.

$8 x - 3 when x = 4$

$y^{3} when y = 5$

Solution

125

$6 a - 2 b when a = 5, b = 7$

$h w when h = 12, w = 3$

Solution

36

Simplify by combining like terms.

ⓐ $6 x + 8 x$
ⓑ $9 m + 10 + m + 3$

In the following exercises, translate each phrase into an algebraic expression.

$5$ more than $x$

Solution

x + 5

the quotient of $12$ and $y$

three times the difference of $a and b$

Solution

3(a − b)

Caroline has $3$ fewer earrings on her left ear than on her right ear. Call the number of earrings on her right ear, $r .$ Write an expression for the number of earrings on her left ear.

In the following exercises, solve each equation.

$n - 6 = 25$

Solution

n = 31

$x + 58 = 71$

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

$15$ less than $y$ is $32 .$

Solution

y − 15 = 32; y = 47

the sum of $a$ and $129$ is $164 .$

List all the multiples of $4$ that are less than $50 .$

Solution

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

Find all the factors of $90 .$

Find the prime factorization of $1080 .$

Solution

2³ ⋅ 3³ ⋅ 5

Find the LCM (Least Common Multiple) of $24$ and $40 .$