Prime Factorisation

Prime Factorisation exercise

• Define the keywords for prime factorisation.

  Hints

  Examples of prime numbers: $2, 3, 5, 7, 11, ...$

  How many factors does each of these numbers have?

  Examples of composite numbers: $4, 6, 8, 9, ...$

  How many factors does each of these numbers have?

  $110 = 2 \times 5 \times 11$ is an example of prime factorisation.

  Solution

  A prime number has only two factors; 1 and itself.

  A composite number has more than two factors.

  Prime factorisation means to write a number as a product of its prime factors.

  A factor tree is a diagram used to find the prime factors of a number.

• Find the prime factorisation for 18.

  Hints

  To start our prime factorisation, we want to find two numbers that multiply to make 18.

  Once we have two numbers, we draw them on to our factor tree. Now check, are they prime?

  If they are prime, we circle them and stop that branch; if they are composite we repeat the first two steps until the ends of each branch are prime factors.

  There are two correct answers.

  Solution

  We can choose any two numbers that multiply to make $18$ to start our prime factorisation, so we have two possible answers here.

  • $18 = 3 \times 6$,

  • $6$ is not prime so we continue,

  • $6 = 3 \times 2 $,

  Putting these together and writing the numbers in numerical order we get $\bf{18 = 2 \times 3 \times 3}$ or $\bf{18 = 2 \times 3^2}$.

  • $18 = 2 \times 9$,

  • $9$ is not prime so we continue,

  • $9 = 3 \times 3$,

  Putting these together and writing the numbers in numerical order we get $\bf{18 = 2 \times 3 \times 3}$ or $\bf{18 = 2 \times 3^2}$

• Complete the prime factorisation for each number.

  Hints

  First, draw a factor tree. Here is the factor tree for 24 to help you.

  The numbers must be in numerical order i.e. smallest to largest.

  Solution

  Here is the factor tree for $44$

  From this we can see that $\bf{44 = 2 \times 2 \times 11}$

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  • $\bf{26 = 2 \times 13}$

  • $\bf{27 = 3 \times 3 \times 3}$

• Work out the prime factorisation for each number.

  Hints

  Use a factor tree to help you, start by finding numbers that multiply to make the starting number.

  For example, the factor tree for $24$ looks like this.

  The circled numbers are your prime factors. Don't forget to write the numbers in numerical order, i.e. from smallest to largest.

  Solution

  In the image we can see the factor tree for $90$

  • $\bf{90 = 2 \times 3 \times 3 \times 5}$

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  • $\bf{28 = 2 \times 2 \times 7}$

  • $\bf{165 = 3 \times 5 \times 11}$

• True or False?

  Hints

  A number is in prime factorisation form if it is written as a product of its prime factors. Product means multiply.

  We continue until all of the factors are prime. The first ten prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$.

  Remember, you can use a factor tree to help you. The image shows a factor tree for $18$.

  First, we used $18 = 3 \times 6$.

  $3$ is prime so circle $3$ and that branch ends there.

  $6$ is not prime so we continue. $6 = 2 \times 3$, these are both prime so we circle those and stop.

  Now we know $18 = 2 \times 3 \times 3$ or in index form $18 = 2 \times 3^2$.

  Be careful, $9$ is not a prime number because $9 = 3 \times 3$.

  There are two correct answers.

  Solution

  $8 = 3 + 5$ is false because this uses addition instead of multiplication.

  $42 = 2 \times 3 \times 7$ is true.

  $35 = 5 \times7$ is true.

  $36 = 2 \times 2 \times 9$ is false because $9$ is not prime.

• What is the prime factorisation of $1{,}260$?

  Hints

  There are many ways to start the factor tree. We can start with $\bf{1{,}260 = 126 \times 10}$ or we can start with $\bf{1{,}260 = 2 \times 630}$.

  Keep going until each branch ends with a prime number i.e. $2, 3, 5, 7, 11$ etc.

  We can write our prime factorisation using exponents (index form).

  $\bf{12 = 2 \times 2 \times 3}$ can also be written as $\bf{12 = 2^2 \times 3}$.

  Solution

  Here are two possible ways to create the factor tree for $1{,}260$. The factor trees may be different but the prime factorisation is the same for any correct factor tree for $1{,}260$.

  $\bf{1{,}260 = 2 \times 2 \times 3 \times 3 \times 5 \times7}$.

  This can also be written as $\bf{1{,}260 = 2^2 \times 3^2 \times 5 \times7}$.