Prime factorisation

Take any whole number greater than $1$. We claim you can write it as a product of primes , and there is essentially only one way to do so. This is one of the most powerful ideas in all of arithmetic.

Concept

The prime factorisation of a number is its expression as a product of prime numbers. For example:

• $12 = 2 \times 2 \times 3 = 2^2 \times 3$
• $30 = 2 \times 3 \times 5$
• $72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$
• $100 = 2 \times 2 \times 5 \times 5 = 2^2 \times 5^2$
• $7 = 7$ (a prime number is its own prime factorisation)

The Fundamental Theorem of Arithmetic says: every whole number greater than $1$ has a prime factorisation, and the factorisation is unique (apart from the order in which the primes are written).

Method 1: Repeated division

The most common method is repeated division by primes:

1. Find the smallest prime that divides $n$. Write it down. Divide.
2. With the new quotient, repeat , smallest prime that divides it.
3. Keep going until the quotient is $1$.

For $84$:

• $84 \div 2 = 42$.
• $42 \div 2 = 21$.
• $21 \div 3 = 7$.
• $7 \div 7 = 1$.

So $84 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7$.

Method 2: Factor tree

A more visual method is the factor tree. Split $n$ into any factor pair, then split each factor that is composite into its own factor pair, and keep going until every branch ends in a prime.

For $60$:

        60
       /  \
      6    10
     / \   / \
    2   3 2   5

Reading the leaves: $60 = 2 \times 3 \times 2 \times 5 = 2^2 \times 3 \times 5$.

It doesn't matter how you start the tree , you always end with the same multiset of primes (by uniqueness).

Using prime factorisation

Once a number is in prime factor form, you can:

• Count its factors. If $n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k}$, then it has $(a_1 + 1)(a_2 + 1) \cdots (a_k + 1)$ factors.
• Find its HCF and LCM with another number , by taking minimum or maximum of each prime's power.
• Test if it is a perfect square, a perfect cube, or so on (all exponents must be multiples of $2$, of $3$, etc.).

Prime factorisation is the secret weapon for many arithmetic questions.

Worked examples

Example 1. Find the prime factorisation of $90$.

• $90 \div 2 = 45$.
• $45 \div 3 = 15$.
• $15 \div 3 = 5$.
• $5 \div 5 = 1$.
• $90 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$.

Example 2. Find the prime factorisation of $144$.

• $144 \div 2 = 72$, $\div 2 = 36$, $\div 2 = 18$, $\div 2 = 9$, $9 \div 3 = 3$, $3 \div 3 = 1$.
• $144 = 2^4 \times 3^2$.

Example 3. How many factors does $72$ have?

• $72 = 2^3 \times 3^2$.
• Number of factors: $(3+1)(2+1) = 4 \times 3 = 12$ factors.
• (List: $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$.)

Example 4. Is $200$ a perfect square?

• $200 = 2^3 \times 5^2$. The power of $2$ is $3$ , odd.
• For a perfect square, every prime power must be even.
• $200$ is not a perfect square.

Example 5. The prime factorisation of a number is $2 \times 3 \times 5 \times 7$. What number is it?

• $2 \times 3 \times 5 \times 7 = 210$.

Try it yourself

1. Find the prime factorisation of $108$.
2. Find the prime factorisation of $225$.
3. How many factors does $48$ have?
4. Is $360$ a perfect square? Is $400$?
5. The prime factorisation of a number is $3^2 \times 5 \times 11$. What is the number?
6. Write the prime factorisation of $1{,}000$.
7. List two numbers whose prime factorisation contains only the prime $2$.
8. Investigate: find a number with exactly $6$ factors. Is there more than one?

Activity

Factor-tree race. Pick a number between $100$ and $200$. Draw a factor tree to find its prime factorisation. Now redraw the tree, but start with a different factor pair at the top. The tree will look different, but the leaves should be the same set of primes. Try this with three different starting splits. This is the "uniqueness" of prime factorisation in action.