Prime Factorisation

(how to write a whole number as a product of its prime factors)


In this section we learn about prime factorisation. In other words we learn how to write a whole number as a product of its prime factors.

Given a whole number \(n\), a prime factor of \(n\) is a factor of \(n\) which is also a prime number.

The first few prime numbers are: \[2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19, \dots \] Every whole number, greater than \(1\), is either one of the prime numbers or can be written as a product of prime numbers. For instance, \(18\) can be written: \[18 = 2\times 3^2\] The fact that this type of factorisation can be done for all whole numbers greater than \(1\) stems from the fundamental theorem of arithmetic.

Fundamental Theorem of Arithmetic

The fundamental theorem of arithmetic states:

Every integer greater than \(1\) is either a prime number or can be written as a product of its prime factors.

This means that every whole number, that is greater than \(1\) can be written as a product of its prime factors (no exceptions). The method for doing this is explained in the following tutorial.

Tutorial: how to write a whole number as a product of its prime factors

Exercise 1

Write each of the following whole numbers as its product of prime factors:

  1. \(36\)
  2. \(90\)
  3. \(196\)
  4. \(900\)
  5. \(384\)
  6. \(3600\)
  7. \(210\)
  8. \(540\)
  9. \(144\)
  10. \(180\)

Note: this exercise can be downloaded as a worksheet to practice with: Worksheet 1

Solution Without Working

We find the following results:

  1. \(36 = 2^2\times 3^2\)
  2. \(90 = 2\times 3^2\times 5\)
  3. \(196 = 2^2\times 7^2\)
  4. \(900 = 2^2\times 3^2 \times 5^2\)
  5. \(384 = 2^7\times 3\)
  6. \(3600 = 2^4 \times 3^2 \times 5^2\)
  7. \(210 = 2\times 3\times 5\times 7\)
  8. \(540 = 2^2 \times 3^3 \times 5\)
  9. \(144 = 2^4 \times 3^2\)
  10. \(180 = 2^2 \times 3^2 \times 5\)


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