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

Revision Note

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Cambridge (CIE) International Maths: ExtendedEdexcel Maths A (Modular): Higher Unit 2Edexcel Maths A: HigherCambridge (CIE) Maths: Extended

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 square root of 2 to the power of 4 cross times 3 squared end root equals 2 squared cross 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 equals 2 to the power of 5 cross times 3 squared cross 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 equals 2 to the power of 4 cross times 2 cross times 3 squared cross times 5

  • Collect the terms with even powers together

    • 1440 equals 2 to the power of 4 cross times 3 squared cross times 2 cross times 5

  • Square root both sides

    • square root of 1440 equals square root of 2 to the power of 4 cross times 3 squared cross times 2 cross times 5 end root

  • Using the rule square root of a b end root equals square root of a square root of b, apply the square root to the terms with the even indices separately to the terms with odd indices

    • square root of 1440 equals square root of 2 to the power of 4 cross times 3 squared end root cross times square root of 2 cross times 5 end root

  • Simplify to find your answer, remembering that square root of a squared b squared end root equals a b

    • square root of 1440 equals 2 squared cross times 3 cross times square root of 10

    • square root of 1440 equals 12 square root of 10

    • 12 square root of 10 is the exact square root of 1440

Worked Example

N equals 2 cubed cross times 3 squared cross times 5 to the power of 7 and A N equals B where Ais 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

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    Author
    Reviewer
    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

    Lucy Kirkham

    Author: Lucy Kirkham

    Expertise: Head of STEM

    Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.