Prime Factors
Every whole number greater than 1 can be written as a product of primes in exactly one way. In NSW Year 8 maths you find that product with a factor tree or repeated division and write it in index form, such as \(300 = 2^2 \times 3 \times 5^2\) (NESA outcome MA4‑IND‑C‑01).
You will use prime factors to find the highest common factor (HCF) and lowest common multiple (LCM), divide large numbers, and find roots such as \(\sqrt{8100}\) and \(\sqrt[3]{512}\) without a calculator. You will also find the smallest number to multiply or divide by to make a square or a cube. Cutting ribbons into equal pieces and tiling the smallest square from rectangles are HCF and LCM problems. It is part of Year 8 Working with Numbers.
Theory
A prime number has exactly two factors, \(1\) and itself. Every whole number greater than \(1\) is a product of primes in exactly one way: \(240 = 2^4 \times 3 \times 5\). This Year 8 maths page shows how to find prime factors and use them for the HCF, the LCM, division and roots.
A prime number can only be divided exactly by \(1\) and itself, such as \(2, 3, 5, 7, 11\). (\(1\) is not prime.)
Repeated division. Divide by the smallest prime that works, over and over, until you reach \(1\). Factor tree: split into any factor pair and keep splitting until every branch ends in a prime. Both give the same primes. Write them in index form, smallest prime first, and check by multiplying back.
HCF (highest common factor): use only the primes that every number has, each with the smaller index. LCM (lowest common multiple): use every prime that appears, each with the larger index.
In problems, the HCF is the largest equal share or longest equal piece; the LCM is the first time two cycles meet again, or the smallest length both numbers fit into exactly.
Dividing takes primes away, so subtract the indices: \(\dfrac{2^5 \times 7^3}{2^2 \times 7} = 2^3 \times 7^2\).
Roots. A square root splits the primes into two equal groups, so halve every index. A cube root divides every index by \(3\). A square number has every index even; a cube number has every index a multiple of \(3\): \(2^4 \times 5^2 = 400 = 20^2\).
Using prime factors (index form):
| To find | Use |
|---|---|
| HCF | shared primes only, smaller index |
| LCM | every prime, larger index |
| a quotient | subtract the indices |
| a square root | halve every index |
| a cube root | divide every index by 3 |
How to write a number as a product of primes
- Divide by \(2\) as many times as you can, then \(3\), then \(5\), and so on.
- Stop when you reach \(1\).
- Write the primes you divided by in index form and check.
How to find the HCF and LCM
- Write each number in index form.
- HCF: shared primes, smaller index. LCM: all primes, larger index.
- Check: the LCM divides by both numbers.
| \(240\) | \(=\) | \(2 \times 120\) |
| \(\) | \(=\) | \(2 \times 2 \times 60\) |
| \(\) | \(=\) | \(2 \times 2 \times 2 \times 30\) |
| \(\) | \(=\) | \(2 \times 2 \times 2 \times 2 \times 15\) |
| \(\) | \(=\) | \(2^4 \times 3 \times 5\) |
\(240 = 2^4 \times 3 \times 5\). Check: \(16 \times 15 = 240\).
| \(42\) | \(=\) | \(2 \times 3 \times 7\) |
| \(63\) | \(=\) | \(3 \times 3 \times 7\) |
The shared primes are \(3\) and \(7\), so the HCF is \(3 \times 7 = 21\).
(ii) Use it to work out \(1500 \div 60\).
(i) \(1500 \div 2 = 750\), \(\div 2 = 375\), \(\div 3 = 125\), \(\div 5 = 25\), \(\div 5 = 5\), \(\div 5 = 1\).
| \(1500\) | \(=\) | \(2^2 \times 3 \times 5^3\) |
| \(\dfrac{2^2 \times 3 \times 5^3}{2^2 \times 3 \times 5}\) | \(=\) | \(5^2 = 25\) |
(i) \(2^2 \times 3 \times 5^3\)
(ii) \(60 = 2^2 \times 3 \times 5\), so \(1500 \div 60 = 25\)
| \(132\) | \(=\) | \(2^2 \times 3 \times 11\) |
| \(180\) | \(=\) | \(2^2 \times 3^2 \times 5\) |
| \(\text{HCF}\) | \(=\) | \(2^2 \times 3 = 12\) |
| \(\text{LCM}\) | \(=\) | \(2^2 \times 3^2 \times 5 \times 11 = 1980\) |
HCF \(= 12\), LCM \(= 1980\). Check: \(1980 \div 132 = 15\) and \(1980 \div 180 = 11\).
(i) \(\sqrt{1296}\)
(ii) \(\sqrt[3]{2744}\)
| \(\sqrt{2^4 \times 3^4}\) | \(=\) | \(2^2 \times 3^2 = 36\) |
| \(\sqrt[3]{2^3 \times 7^3}\) | \(=\) | \(2 \times 7 = 14\) |
(i) \(1296 = 16 \times 81 = 2^4 \times 3^4\), so \(36\)
(ii) \(2744 = 8 \times 343 = 2^3 \times 7^3\), so \(14\)
(i) What is the smallest whole number that multiplies \(675\) to a cube number?
(ii) What is the smallest whole number you can divide \(675\) by to leave a square number?
| \(675 \times 5\) | \(=\) | \(3^3 \times 5^3 = 15^3\) |
| \(675 \div 3\) | \(=\) | \(3^2 \times 5^2 = 15^2\) |
(i) \(5\) (the \(5^2\) needs one more \(5\)): \(3375 = 15^3\)
(ii) \(3\) (the \(3^3\) has an odd index): \(225 = 15^2\)
Common pitfalls
Frequently asked questions
What is a prime number?
A whole number greater than 1 whose only factors are 1 and itself, such as 2, 3, 5, 7 and 11.
How do you write a number as a product of prime factors?
Divide by the smallest prime that works, again and again, until you reach 1: \(60 = 2^2 \times 3 \times 5\).
How do you find the HCF using prime factors?
Multiply the primes both numbers share, using the smaller index of each.
How do you find the LCM using prime factors?
Multiply every prime that appears, using the larger index of each.
How can prime factors help find a square root?
Halve every index: \(\sqrt{2^4 \times 3^2} = 2^2 \times 3 = 12\).