1. Number

Middle Years

Lesson

All whole numbers except for $1$1 are either prime or composite, and composite numbers can always be written as a product of primes. Finding this product (called a prime factorisation) can be very useful.

Primes and composites

A number is prime if it has exactly two factors: $1$1, and itself.

A number is composite if it has more than two factors.

Learn more about prime numbers in our investigation: The Sieve of Eratosthenes.

One of the best ways to find a prime factorisation is by using a factor tree. We start with the number we want to investigate, draw a box around it, and draw two lines coming out of it. Here is how we might start with the number $12$12:

We then put two numbers that multiply to make $12$12, such as $4$4 and $3$3, at the end of each of the lines:

Because $3$3 is a prime number, we circle it. Since $4$4 is not a prime number, we draw a box around it instead:

We then repeat the process with $4$4, which is $2\times2$2×2:

And since $2$2 is prime, we circle both of these numbers:

This is a completed factor tree for $12$12, and it tells us that $12=2\times2\times3$12=2×2×3. Multiplying the circled numbers at the end of each branch together always makes the original number.

Factor trees are not always unique - here is another factor tree for $12$12:

Even though the number in the box is different, the numbers at the end of the branches will always be the same for any number - they will just be in a different order.

Here is a factor tree for $360$360:

We can therefore write:

$360=3\times5\times2\times2\times2\times3$360=3×5×2×2×2×3

We usually rewrite the factors so they are in ascending order, like this:

$360=2\times2\times2\times3\times3\times5$360=2×2×2×3×3×5

We can use index form to make the expression shorter as well:

$360=2^3\times3^2\times5$360=23×32×5

Notice that the factor tree for $12$12 we made earlier is a smaller part of the factor tree for $360$360. This is because $12$12 is a factor of $360$360, and when we write $360=2\times2\times2\times3\times3\times5$360=2×2×2×3×3×5 we can recognise the prime factorisation of $12$12 inside it: $360=2\times\left(2\times2\times3\right)\times3\times5$360=2×(2×2×3)×3×5.

Factor trees are useful because every number we write as we make it is a factor of the original number. We don't always see every factor appear, though - for example, $9$9 is a factor of $360$360, but it does not appear in the tree above.

To find every factor of a number we need to combine the prime factors in every possible way. First, we find the prime factorisation like we did before, such as:

$12=2\times2\times3$12=2×2×3

We then combine these factors in every possible way. Every factor of $12$12 can have no $2$2s, one $2$2, or two $2$2s in its prime factorisation. Similarly, every factor of $12$12 can have no $3$3s, or one $3$3. Here we draw this out in a table:

Factors of $12$12 | |||
---|---|---|---|

No $2$2 | One $2$2 | Two $2$2s | |

No $3$3 | $1$1 | $2$2 | $2\times2$2×2 |

One $3$3 | $3$3 | $2\times3$2×3 | $2\times2\times3$2×2×3 |

We then perform each of the multiplications to find all the factors:

Factors of $12$12 | |||
---|---|---|---|

No $2$2 | One $2$2 | Two $2$2s | |

No $3$3 | $1$1 | $2$2 | $4$4 |

One $3$3 | $3$3 | $6$6 | $12$12 |

The factors of $12$12 are $1$1, $2$2, $3$3, $4$4, $6$6, and $12$12.

Here is how we can do it for $360$360 - every factor either has $5$5 as a factor or it doesn't, it has between zero and two $3$3s, and between zero and three $2$2s.

Factors of $360$360 | |||||
---|---|---|---|---|---|

No $2$2 | One $2$2 | Two $2$2s | Three $2$2s | ||

No $5$5 | No $3$3 | $1$1 | $2$2 | $2\times2$2×2 | $2\times2\times2$2×2×2 |

One $3$3 | $3$3 | $2\times3$2×3 | $2\times2\times3$2×2×3 | $2\times2\times2\times3$2×2×2×3 | |

Two $3$3s | $3\times3$3×3 | $2\times3\times3$2×3×3 | $2\times2\times3\times3$2×2×3×3 | $2\times2\times2\times3\times3$2×2×2×3×3 | |

One $5$5 | No $3$3 | $5$5 | $2\times5$2×5 | $2\times2\times5$2×2×5 | $2\times2\times2\times5$2×2×2×5 |

One $3$3 | $3\times5$3×5 | $2\times3\times5$2×3×5 | $2\times2\times3\times5$2×2×3×5 | $2\times2\times2\times3\times5$2×2×2×3×5 | |

Two $3$3s | $3\times3\times5$3×3×5 | $2\times3\times3\times5$2×3×3×5 | $2\times2\times3\times3\times5$2×2×3×3×5 | $2\times2\times2\times3\times3\times5$2×2×2×3×3×5 |

This table shows all the possible ways to multiply the prime factors together. We evaluate the multiplications to find all the factors:

Factors of $360$360 | |||||
---|---|---|---|---|---|

No $2$2 | One $2$2 | Two $2$2s | Three $2$2s | ||

No $5$5 | No $3$3 | $1$1 | $2$2 | $4$4 | $8$8 |

One $3$3 | $3$3 | $6$6 | $12$12 | $24$24 | |

Two $3$3s | $9$9 | $18$18 | $36$36 | $72$72 | |

One $5$5 | No $3$3 | $5$5 | $10$10 | $20$20 | $40$40 |

One $3$3 | $15$15 | $30$30 | $60$60 | $120$120 | |

Two $3$3s | $45$45 | $90$90 | $180$180 | $360$360 |

The factors of $360$360 are $1$1, $2$2, $3$3, $4$4, $5$5, $6$6, $8$8, $9$9, $10$10, $12$12, $15$15, $18$18, $20$20, $24$24, $30$30, $36$36, $40$40, $45$45, $60$60, $72$72, $90$90, $120$120, $180$180, and $360$360.

A number has the following factor tree:

What is this number at the top of the tree?

Write $148$148 as a product of prime factors in expanded form.

In this question we will be finding the factors of $20$20.

First, write $20$20 as a product of prime factors in expanded form.

Using your answer from part (a), list all the factors of $20$20, separated by commas.