Uses of Prime Factor Decomposition (Cambridge (CIE) IGCSE International Maths)

Revision Note

Written by: Dan Finlay

Uses of Prime Factor Decomposition

When a number has been written as its prime factor decomposition (PFD), it can be used to find out if that number is a square or cube number, or to find the square root of that number without a calculator.

How can I use PFD to identify a square or cube number?

If all the indices in the prime factor decomposition of a number are even, then that number is a square number
- E.g. The prime factor decomposition of 7056 is 2⁴ × 3² × 7²
- All powers are even so it must be a square number
  - It can be written as (2² × 3 × 7)²
If all the indices in the prime factor decomposition of a number are multiples of 3, then that number is a cube number
- E.g. The prime factor decomposition of 1728000 is 2⁹ × 3³ × 5³
- All powers are multiples of 3 so it must be a cube number
  - It can be written as (2³ × 3 × 5)³

How can I use PFD to find the square root of a square number?

Write the number in its prime factor decomposition
- All the indices should be even if it is a square number
For example, to find the square root of 144 = 2⁴ × 3²
- Halve all of the indices
  - 2² × 3
  - So $\sqrt{2^{4} \times 3^{2}} = 2^{2} \times 3$
This is the prime factor decomposition of the square root of the number
- To find it as an integer, multiply the prime factors together
- 2² × 3 = 12, so the square root of 144 is 12

How can I use PFD to find the exact square root of a number?

If the number is not a square number, its exact square root can still be found using its prime factor decomposition
Write the number in its prime factor decomposition
- $1440 = 2^{5} \times 3^{2} \times 5$
Rewrite the prime factor decomposition with as many even indices as you can
- E.g. 2³ = 2² × 2, or 5⁷ = 5⁶ × 5
- $1440 = 2^{4} \times 2 \times 3^{2} \times 5$
Collect the terms with even powers together
- $1440 = 2^{4} \times 3^{2} \times 2 \times 5$
Square root both sides
- $\sqrt{1440} = \sqrt{2^{4} \times 3^{2} \times 2 \times 5}$
Using the rule $\sqrt{a b} = \sqrt{a} \sqrt{b}$ , apply the square root to the terms with the even indices separately to the terms with odd indices
- $\sqrt{1440} = \sqrt{2^{4} \times 3^{2}} \times \sqrt{2 \times 5}$
Simplify to find your answer, remembering that $\sqrt{a^{2} b^{2}} = a b$
- $\sqrt{1440} = 2^{2} \times 3 \times \sqrt{10}$
- $\sqrt{1440} = 12 \sqrt{10}$
- $12 \sqrt{10}$ is the exact square root of $1440$

Worked Example

$N = 2^{3} \times 3^{2} \times 5^{7}$ and $A N = B$ where $A$ is an integer and $B$ is a non-zero square number.

Find the smallest value of $A$ .

Substitute N = 2³× 3²× 5⁷ into the formula AN = B

A(2³× 3²× 5⁷ ) = B

To be a square number, the prime factors of AN must all have even powers

Consider the prime factors A needs to have to make all the values on the left hand side have even powers

(2 × 5) (2³× 3²× 5⁷) = B

2⁴× 3²× 5⁸= B

So A, when written as a product of its prime factors, is 2 × 5

Make sure you write A as an integer value in the answer

A = 10

Last updated: 23 May 2024

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