Is [katex] \bf{1} [/katex] a prime number?

No, [katex] 1 [/katex] is not a prime number because in order for a number to be prime it must be greater than [katex] 1. [/katex]Also, prime numbers have exactly [katex] 2 [/katex] factors and [katex] 1 [/katex] only has one factor.

Prime Factors - Steps, Examples & Questions

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Prime factors

Here is everything you need to know about prime factors. You will learn the fundamental theorem of arithmetic, how to express a number as a product of prime factors and use the product of prime factors to recognize special numbers such as square numbers and cube numbers.

What are prime factors?

Prime factors are prime numbers that are factors of another number.

A composite number is the product of two or more factors. All of these types of numbers are integers (whole numbers).

For example,

In number theory, the Fundamental Theorem of Arithmetic states that every integer greater than one is a prime number or can be represented by a product of prime numbers.

For example,

This would mean that the number $\bf{240}$ is a composite number with the prime factors of $\bf{2, 3}$ and $\bf{5}$ .

Expressing a composite number as a product of prime factors can be utilized for a wide variety of problems such as:

calculating the greatest common factor (GCF)
calculating the least common multiple (LCM)
simplifying square roots (radicals)
determining whether a number is a square number or cube number
factorization
calculating the square roots of numbers

and much more.

We can use prime factors in order to recognize different types of numbers from their exponents.

Number Property	Description	Example
Square Numbers	The exponent of each prime factor is even	$3^2= 9$ $3^4=81$
Cube Numbers	The exponent of each prime factor is a multiple of $3$	$3^3= 27$ $3^6=729$
Even Number	$2$ is a prime factor	$4 = 2 \times 2$ $6 = 3 \times 2$

To find the prime factors of a number, we need to continue to divide the composite number by prime numbers, until we are left with just prime factors.

Looking back at the example of $6=2\times{3},$ we can say,

Here, the divisor is $\bf{3}$ as we are dividing $6$ by $3$ to get a quotient of $2.$

The divisor will always be a prime number for these questions and if the quotient is a prime number, we have found the full decomposition.

What are prime factors?

Common Core State Standards

How does this relate to 6th grade math?

Grade 6 – The Number System (6.NS.2)
Fluently divide multi-digit numbers using the standard algorithm.

Grade 6 – Expressions and Equations (6.EE.1)
Write and evaluate numerical expressions involving whole-number exponents.

Product of prime factors

Any positive integer can be written as a product of its prime factors. This means that we can take any positive number and write it as a series of prime numbers being multiplied.

For example,

$6$ is a product of $2$ and $3,$ so can be written as $2\times{3}=6$

$9$ is a product of $3$ and $3,$ so can be written as $3\times{3}=9$

We can use a process called prime factorization using prime factor trees in order to work out the product of prime factors.

For example,

Write $36$ as a product of prime factors.

36\div{2}=18

18\div{2}=9

9\div{3}=3

So,

36=2\times{2}\times{3}\times{3}

This can also be written using exponents.

36=2^{2}\times3^{2}

Let’s look at another example!

Write $54$ as a product of prime factors.

54\div{2}=27

27\div{3}=9

9\div{3}=3

So,

54=2\times{3}\times{3}\times{3}

Writing using exponents,

54=2\times3^{3}

Divisibility rules

When finding distinct prime factors it is useful to use the divisibility rules.

$\hspace{2.9cm}$ Divisibility Rules!

$\hspace{2.3cm}$ A number is divisible by…….


$2$		if the last digit is even or zero
$3$		if the sum of the digits is divisible by three
$4$		if the last two digits are divisible by four
$5$		if the last digit is zero or five
$6$		if the number is divisible by both two and three
$8$		if the last three digits are divisible by eight
$9$		if the sum of the digits is divisible by nine
$10$		if the last digit is zero

How to find prime factors of a composite number

In order to find prime factors of a composite number:

Divide the composite number by a suitable prime number.
Divide the quotient by another suitable prime number.
Continue until the final quotient is a prime number, then state the solution.

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Prime factors examples

Example 1: composite number <20

What are the prime factors of $18?$

Divide the composite number by a suitable prime number.

As $18$ is an even number, we can divide $18$ by $2.$

18\div{2}=9

The first prime factor of $18$ is $2.$

$9$ is not a prime number so we continue to Step 2.

2Divide the quotient by another suitable prime number.

$9$ is a multiple of $3$ and so we can divide $9$ by $3.$

9\div{3}=3

The second prime factor is $3.$

$3$ is a prime number so the third prime factor is $3.$

3Continue until the final quotient is a prime number, then state the solution.

As factors are multiplied together to achieve the composite number, we can say

18=2\times{3}\times{3}

(or in exponent form)

18=2\times3^{2}.

The final solution is therefore

18=2\times3^{2}.

Example 2: composite number <50

Show that $42=2\times{3}\times{7}.$

Divide the composite number by a suitable prime number.

As $42$ is an even number, we can divide $42$ by $2.$

$42\div{2}=21$

The first prime factor of $42$ is $2.$

$21$ is not a prime number so we continue to Step 2.

Divide the quotient by another suitable prime number.

$21$ is a multiple of $3$ and so we can divide $21$ by $3.$

$21\div{3}=7$

The second prime factor is $3.$

$7$ is a prime number so the third prime factor is $7.$

Continue until the final quotient is a prime number, then state the solution.

As factors are multiplied together to achieve the composite number, we can say

$42=2\times{3}\times{7}$

An exponent form does not occur for this example as there are no repeated factors that we can simplify.

The final solution is therefore

$42=2\times{3}\times{7}.$

Example 3: composite number <100

Express $75$ as a product of prime factors. Write your answer in index form.

Divide the composite number by a suitable prime number.

As $75$ ends in a $5,$ we can divide $75$ by $5.$

$75\div{5}=15$

The first prime factor of $75$ is $5.$

$15$ is not a prime number so we continue to Step 2.

Divide the quotient by another suitable prime number.

$15$ also ends in a $5$ and so we can divide $15$ by $5.$

$15\div{5}=3$

The second prime factor is $5.$

$3$ is a prime number so the third prime factor is $3.$

Continue until the final quotient is a prime number, then state the solution.

As factors are multiplied together to achieve the composite number, we can say

$75=3\times{5}\times{5}$

$75=3\times5^{2}$

The final solution in index form is therefore

$75=3\times5^{2}.$

Example 4: composite number <500

Express $462$ as a product of prime factors.

Divide the composite number by a suitable prime number.

As $462$ is an even number, we can divide $462$ by $2.$

$462\div{2}=231$

The first prime factor of $462$ is $2.$

$231$ is not a prime number so we continue to Step 2.

Divide the quotient by another suitable prime number.

The sum of the digits of $231 \; (2+3+1)=6$ which is a multiple of $3$ and so we can divide $231$ by $3.$

$231\div{3}=77$

The second prime factor is $3.$

$77$ is not a prime number so we need to continue to divide by prime numbers until the final quotient is prime.

Continue until the final quotient is a prime number, then state the solution.

As $77$ is a multiple of $7,$ we can divide $77$ by $7.$

$77\div{7}=11$

The third prime factor is $7.$

$11$ is a prime number so the fourth prime factor is $11.$

As factors are multiplied together to achieve the composite number, we can say

$462=2\times{3}\times{7}\times{11}$

The final solution in index form is therefore

$462=2\times{3}\times{7}\times{11}.$

Example 5: composite number <1000

Show that $900$ is a square number.

Divide the composite number by a suitable prime number.

As the sum of the digits of $900 \; (9+0+0)=9,$ we can divide $900$ by $3.$

$900\div{3}=300$

The first prime factor of $900$ is $3.$

$300$ is not a prime number so we continue to Step 2.

Divide the quotient by another suitable prime number.

The sum of the digits of $300 \; (3+0+0)=3$ which is a multiple of $3$ and so we can divide $300$ by $3.$

$300\div{3}=100$

The second prime factor is $3.$

$100$ is not a prime number so we need to continue to divide by prime numbers until the final quotient is prime.

Continue until the final quotient is a prime number, then state the solution.

As $100$ is an even number, we can divide $100$ by $2.$

$100\div{2}=50$

The third prime factor is $2.$

$50$ is an even number so we can divide $50$ by $2.$

$50\div{2}=25$

The fourth prime factor is $2.$

$25$ ends in a $5$ so we can divide $25$ by $5.$

$25\div{5}=5$

The final two prime factors are $5$ and $5.$

As factors are multiplied together to achieve the composite number, we can say

$900=2\times2\times3\times3\times5\times5$

$900=(2\times3\times5)\times(2\times3\times5)$

$900=(2\times3\times5)^2$

So $900$ is a square number.

Example 6: simplifying fractions

Simplify $\cfrac{104}{124}$ using prime factors.

Divide the numerator by a suitable prime number.

$104$ is an even number so we can divide it by $2.$

$104\div{2}=52$

The first prime factor of $104$ is $2.$

$52$ is not a prime number.

Divide the quotient by another suitable prime number.

$52$ ends in an even number so it is divisible by $2.$

$52\div{2}=26$

$2$ is another prime factor.

$26$ is not a prime number.

Continue until the final quotient is a prime number.

$26$ ends in an even number so it is divisible by $2.$

$26\div{2}=13$

$2$ is another prime factor.

$13$ is also a prime factor.

Rewrite the numerator as prime factors multiplied together.

\cfrac{2\times2\times2\times13}{124}

Do the same process for the denominator.

The last digit of $124$ is even so it is divisible by $2.$

$124\div{2}=62$

$2$ is a prime factor.

$62$ is not prime.

Continue until the final quotient is a prime number.

The last digit of $62$ is even so it is divisible by $2.$

$62\div{2}=31$

$2$ is another prime factor.

$31$ is a prime factor.

Rewrite the denominator as prime factors multiplied together.

\cfrac{2\times2\times2\times13}{2\times2\times31}

Look for matching common prime factors in the numerator and the denominator.

Two of the $2’s$ in the numerator match with two of the $2’s$ in the denominator. We can simplify the fraction by crossing out the matching factors.

Rewrite the fraction by multiplying the factors together.

\cfrac{2\times 13}{31}=\cfrac{26}{31}

Teaching tips for finding prime factors

Reinforce visual models to help students conceptualize the concept.

Start by dividing the given number by the smallest prime number.

Then using that same strategy, divide that quotient by the smallest prime number.

Keep repeating the process, until the quotient becomes $1.$

Lastly, multiply all the prime factors together.

Expose students to cryptography, the study of secret codes, which relies heavily on prime numbers.

Easy mistakes to make

Listing factors instead of writing them as a product (multiplying)
Given example 1, the solution would be incorrectly written as $18=2,3,3$ instead of $18=2\times{3}\times{3}.$ Factors are multiplied by one another and so the solution should contain the multiplication symbol.

Thinking all odd numbers are prime numbers

Thinking $\bf{1}$ is a prime number

Incorrect knowledge of prime numbers
It is easy to find a factor of a composite number and not specifically a prime factor. Take example 6 where we were dividing $900$ by $3.$ A more obvious start could have been to divide by $9,$ however $9$ is not a prime number. When we look at larger composite numbers, there is a much more clear and efficient method to express the prime factors of a number.

Practice prime factors questions

4\times{4}

2^{4}

1,2,4,8,16

2\times{2}\times{4}

16\div{2}=8

8\div{2}=4

4\div{2}=2

$2$ is prime so we have found the prime factors.

16=2\times{2}\times{2}\times{2}=2^{4}

1\times3^{2}\times7

2\times3\times7

3\times21

3^{2}\times7

63\div3=21

21\div3=7

$7$ is prime.

63=3\times3\times7=3^{2}\times7

2\times3\times41

2^{3}\times3\times7

2^{3}\times5\times7

2^{3}\times3^{3}

246\div2=123

123\div3=41

$41$ is prime.

246=2\times3\times41

\cfrac{66}{147}

\cfrac{44}{98}

\cfrac{11}{14}

\cfrac{22}{49}

Divide the numerator by a suitable prime number. Since $198$ ends in an even number, divide it by $2.$

198\div2=99

$99$ is not a prime number, so continue to divide until all the factors are prime.

The sum of the digits of $99$ is divisible by $3.$

99\div3=33

The sum of the digits of $33$ is divisible by $3.$

33\div3=11

Now do the same thing with the denominator. Divide the denominator by a suitable prime number. The sum of the digits of $441$ is divisible by $3.$

441\div3=147

The sum of the digits of $147$ is divisible by $3.$

147\div3=49

$49$ is a multiple of $7.$

49\div7=7

Write the numerator and denominator as a product of prime numbers. Match the common prime factors that are the same in order to simplify.

US Web Page Prime Factors image 13

Multiply the factors in the numerator together and the factors in the denominator together.

\cfrac{2 \times 11}{7 \times 7}=\cfrac{22}{49}

2^{2}\times9

2^{2}\times3^{2}

2\times{2}\times{3}\times{3}

They all show $36$ as a product of its prime factors

Although $2^{2}\times9=36,$ $9$ is not a prime number.

a=3, \; b=2, \; c=5

a=4, \; b=2, \; c=5

a=4, \; b=2, \; c=7

a=2, \; b=3, \; c=5

720 \div 2 =360

360 \div 2 =180

180\div2=90

90\div2=45

45\div3=15

15\div3=5

$5$ is prime.

720=2^{4}\times3^{2}\times 5

a=4, \; b=2, \; c=5

Prime factors FAQs

Is $\bf{1}$ a prime number?

No, $1$ is not a prime number because in order for a number to be prime it must be greater than $1.$ Also, prime numbers have exactly $2$ factors and $1$ only has one factor.

Can a composite number have more than one of the same prime factors?

Yes, a composite number can have the same prime factors more than once.
For example, $8=2 \times2\times2$ .

Are only odd numbers considered prime factors?

Most prime factors are odd, but not all odd numbers are prime. For example $2$ is a prime number and it’s even. $9$ is an odd number and it’s not prime.

Does prime factorization and the factor tree method mean the same thing?

The factor tree method is a strategy for finding the prime factorization of a number.

How does prime factors apply to the real world?

Computer coders use prime factorization of numbers to code.

Why is finding prime factors important?

It helps us to break down large numbers.

The next lessons are

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