C1.1; E1.1 - Prime factorisation | Cambridge IGCSE Mathematics

Different factor pairs can lead to the same set of prime building blocks. You will break a positive integer into prime factors using a tree or repeated division, then write repeated factors compactly using powers.

Prime factors from a factor tree

If one person starts $72$ as $8\times9$ and another starts it as $6\times12$ , will they finish with different prime factors? The starting choices differ, but both routes must preserve the product at every step.

You already know

A prime number has exactly two distinct positive factors. A factor pair multiplies to the original number. For example, $8\times9=72$ , but neither $8$ nor $9$ is prime.

A factor tree splits a number into a factor pair. Each composite factor splits again; a prime is a stopping point. Use pairs greater than $1$ , so that each split makes the numbers smaller. The final numbers at the branch ends multiply to the starting number because every split replaces a number by an equal product.

Product of prime factors

A product of prime factors writes an integer greater than $1$ as prime numbers multiplied together. Every factor in the final product must be prime. For example, $18=2\times3\times3$ ; $18=2\times9$ is an equal product, but is not yet a product of prime factors.

[DIAGRAM: asset_name: Two factor trees for seventy-two; asset_slug: two_factor_trees; caption: Different first splits give the same prime factors, including repeated copies.; description: Two exact factor trees for seventy-two. The left splits into eight and nine, then eight into two and four, four into two and two, and nine into three and three. The right splits into six and twelve, six into two and three, twelve into three and four, and four into two and two. Prime branch ends are highlighted in pale green. Each tree yields three twos and two threes.]
Different first splits give the same prime factors, including repeated copies.

Worked example 1: starting with eight times nine

Question. Express $72$ as a product of prime factors, starting with $8\times9$ .

Plan. Split each composite branch until every branch end is prime.

\begin{aligned} 72&=8\times9 && \text{the chosen first split}\\ &=(2\times4)\times(3\times3) && \text{split both composites}\\ &=2\times2\times2\times3\times3 && \text{split 4; all ends are prime} \end{aligned}

Answer. $72=2\times2\times2\times3\times3$ .

Check. $2\times2\times2=8$ and $3\times3=9$ , so the final product is $8\times9=72$ . Each of $2$ and $3$ is prime.

Keep $72$ and change only the first split. This checks whether the result depends on your starting choice.

Worked example 2: starting with six times twelve

Question. Express $72$ as a product of prime factors, starting with $6\times12$ .

Plan. Split $6$ and $12$ , and continue until only prime branch ends remain.

\begin{aligned} 72&=6\times12 && \text{a different first split}\\ &=(2\times3)\times(3\times4) && \text{split both composites}\\ &=2\times3\times3\times2\times2 && \text{split the remaining 4}\\ &=2\times2\times2\times3\times3 && \text{reorder the factors} \end{aligned}

Answer. $72=2\times2\times2\times3\times3$ , the same prime factors as before.

Check. Group the factors another way: $(2\times3)\times(2\times3)\times2=6\times6\times2=72$ .

This agreement is a general property: for an integer greater than $1$ , a complete prime factorisation is unique apart from the order of its factors. A prime such as $17$ already has a one-factor prime factorisation, $17$ . The number $1$ has no prime factors; do not put $1$ at the end of a prime-factor product.

Watch out

Do not multiply every number written in the tree. Intermediate numbers such as $8$ , $9$ and $4$ have already been replaced by their branches. Read only the prime branch ends, including every repeated copy.

Repeated division into primes

You can find prime factors without drawing a branching tree. Divide by a prime factor, keep that prime, then divide the quotient again. Stop when the quotient becomes $1$ . This builds a single chain of exact divisions.

Start with the smallest suitable prime. If $2$ divides the current number exactly, use it and test $2$ again. Once $2$ no longer divides, try $3$ , then $5$ , then the next primes as needed. Skipping $4$ is safe: any number divisible by $4$ would already be divisible by $2$ .

Worked example 3: keeping a divisor while it still works

Question. Express $84$ as a product of prime factors using repeated division.

Plan. Divide by the smallest prime available at each stage, recording every divisor.

\begin{aligned} 84\div2&=42 && \text{record one factor of 2}\\ 42\div2&=21 && \text{2 still divides exactly}\\ 21\div3&=7 && \text{21 is odd; try 3}\\ 7\div7&=1 && \text{7 is prime; finish} \end{aligned}

Answer. $84=2\times2\times3\times7$ .

Check. Rebuild in reverse: $1\times7=7$ , then $7\times3=21$ , $21\times2=42$ , and $42\times2=84$ .

Change the starting value to $126$ . The first quotient is now odd, so keeping $2$ as the next divisor would fail.

Worked example 4: moving to the next prime

Question. Express $126$ as a product of prime factors using repeated division.

Plan. Use $2$ once, then choose primes that divide each new quotient exactly.

\begin{aligned} 126\div2&=63 && \text{record 2}\\ 63\div3&=21 && \text{63 is odd; record 3}\\ 21\div3&=7 && \text{3 still divides exactly}\\ 7\div7&=1 && \text{record 7 and finish} \end{aligned}

Answer. $126=2\times3\times3\times7$ .

Check. $2\times7=14$ and $3\times3=9$ , giving $14\times9=126$ .

Why keep the divisors? Each line says the number before division equals the prime divisor times the next number. Reading back through the chain multiplies all the saved primes by the final $1$ , reconstructing the original integer.

Watch out

A quotient in the middle of the chain need not be prime. The divisors you record must be prime. For example, dividing $84$ by $4$ is exact, but you would still need to replace $4$ by $2\times2$ .

Writing repeated prime factors as powers

Repeated factors can be shortened without changing their meaning. For example, $2\times2\times2$ can be written $2^3$ . The small raised $3$ counts three factors of $2$ ; it does not mean $2\times3$ .

Power notation

In a power such as $2^3$ , the base is $2$ and the index is $3$ . A positive whole-number index tells you how many copies of the base are multiplied together. Thus $2^3=2\times2\times2=8$ , and $3^2=3\times3=9$ .

For a prime factorisation, collect equal primes together and count their copies. A prime used once is normally written without an index: write $5$ , rather than $5^1$ . This is only a compact record of repeated multiplication; you do not need rules for combining general powers yet.

Worked example 5: counting repeated copies

Question. Write $2\times2\times2\times3\times3$ using powers, and work out its value.

Plan. Count the copies of each prime, then evaluate each repeated product.

\begin{aligned} 2\times2\times2\times3\times3&=2^3\times3^2 && \text{three twos and two threes}\\ &=8\times9 && \text{evaluate the powers}\\ &=72 && \text{multiply} \end{aligned}

Answer. $2^3\times3^2=72$ .

Check. Expand the powers again: $(2\times2\times2)(3\times3)$ . There are exactly the original five prime factors.

Replace one factor of $2$ by a factor of $3$ . The number of factors stays five, but the two indices change.

Worked example 6: changing one prime factor

Question. Write $2\times2\times3\times3\times3$ using powers, and work out its value.

Plan. Recount each prime instead of keeping the indices from the previous example.

\begin{aligned} 2\times2\times3\times3\times3&=2^2\times3^3 && \text{two twos and three threes}\\ &=4\times27 && \text{evaluate the powers}\\ &=108 && \text{multiply} \end{aligned}

Answer. $2^2\times3^3=108$ .

Check. Replacing one $2$ in the previous product by $3$ divides its value by $2$ , then multiplies by $3$ : $72\div2\times3=36\times3=108$ .

The opening factor trees both finish at $2^3\times3^2$ . You can choose the first split to make the arithmetic convenient, then use the prime product to check the result. In the next lesson, keeping track of repeated copies will help you find common factors and common multiples.

Watch out

$2^3\ne2\times3$ : the left side is $8$ , while the right side is $6$ . The index counts factors, not equal groups being added.

Quick Recap

Split composite factors until every branch end is prime.
Different valid factor trees give the same prime factors, apart from their order.
Repeated division records the prime divisors and ends at $1$ .
A positive whole-number index counts repeated factors: $2^3$ means three factors of $2$ .

The two starting routes for $72$ agree: each gives $2\times2\times2\times3\times3=2^3\times3^2$ .

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C1.1; E1.1 - Prime factorisation

Prime factors from a factor tree

Repeated division into primes

Writing repeated prime factors as powers

Quick Recap

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