Year 7 - Powers and prime factors - Free practice

Year 7 Powers and Prime Factors Worksheets & Practice

From expanded notation to prime factorisation. Worked solutions every time, free.

Quick answer

What are powers and prime factors in Year 7 maths?

Every natural number can be written as a product of prime numbers using exponent notation. Find the prime factors with a factor tree or repeated division, then group repeated factors as powers. For example, 72 = 2³ × 3² because 72 = 2 × 2 × 2 × 3 × 3.

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Six questions across Foundation, Standard and Extension, with full worked solutions, written to the Australian Curriculum.

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Worked examples

Year 7 powers and prime factors: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Prime factorisation using a factor tree
Express $72$ as a product of prime factors using exponent notation.
Show worked solution
Key idea: A factor tree splits a number into factor pairs, working down until every branch ends in a prime number. Collect repeated primes using exponent notation.
1. Split $72$ into $8 \times 9$ .
2. Split $8$ into $2 \times 4$ , then $4$ into $2 \times 2$ . So $8 = 2 \times 2 \times 2$ .
3. Split $9$ into $3 \times 3$ .
4. Collect all prime factors: 2, 2, 2, 3, 3.
5. Write using exponents:
  $72 = 2^3 \times 3^2$ Answer: $72 = 2^3 \times 3^2$
Example 2 · Prime factorisation using repeated division
Express $360$ as a product of prime factors using exponent notation. Use repeated division to find the prime factors.
Show worked solution
Key idea: Divide repeatedly by the smallest prime that goes in exactly, starting with $2$ , then $3$ , then $5$ , and so on, until the result is $1$ .
1. $360 \div 2 = 180$
2. $180 \div 2 = 90$
3. $90 \div 2 = 45$
4. $45 \div 3 = 15$
5. $15 \div 3 = 5$
6. $5 \div 5 = 1$
Prime factors used: 2, 2, 2, 3, 3, 5. Writing with exponents:
$360 = 2^3 \times 3^2 \times 5$ Answer: $360 = 2^3 \times 3^2 \times 5$
Example 3 · Prime factorisation in a real-life context
A school canteen orders $180$ muesli bars to pack into identical bags. The canteen manager wants every bag to contain the same number of bars, and also wants to know every possible bag size that divides $180$ exactly. To find these, she first writes $180$ as a product of prime factors. Express $180$ as a product of prime factors using exponent notation, then use your answer to state how many factors $180$ has in total.
Show worked solution
Key idea: The prime factorisation of a number lets you count its total number of factors using the exponent rule: if $n = p^a \times q^b \times \ldots$ , then the number of factors is $(a+1)(b+1)\ldots$
1. Divide repeatedly by primes:
  $\begin{aligned} 180 \div 2 &= 90 \\ 90 \div 2 &= 45 \\ 45 \div 3 &= 15 \\ 15 \div 3 &= 5 \\ 5 \div 5 &= 1 \end{aligned}$
2. Write using exponents: $180 = 2^2 \times 3^2 \times 5$ .
3. Count the factors: $(2+1)(2+1)(1+1) = 3 \times 3 \times 2 = 18$ .
4. So there are $18$ possible bag sizes (for example 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180 bars per bag).
Answer: $180 = 2^2 \times 3^2 \times 5$ ; there are $18$ possible bag sizes.

Common mistakes

Where students usually lose marks

Stopping the factor tree too early

Student writes $72 = 8 \times 9$ , then stops because $8$ and $9$ look like small numbers.

Keep splitting every branch until every number at the end is prime. $8$ and $9$ are not prime ( $8 = 2^3$ and $9 = 3^2$ ), so they must be split further.

Including 1 as a prime factor

Student writes $72 = 1 \times 2^3 \times 3^2$ and counts $1$ as a prime.

By definition, $1$ is neither prime nor composite. Prime factorisation only uses primes ( $2, 3, 5, 7, \ldots$ ), so $1$ is never listed as a factor.

Confusing the base and the exponent

Student reads $2^3$ as $2 \times 3 = 6$ instead of $2 \times 2 \times 2 = 8$ .

The exponent (the small raised number) tells you how many times to multiply the base by itself, not what to multiply it by. So $2^3 = 2 \times 2 \times 2 = 8$ .

Not dividing by the smallest prime first

Student divides $360 \div 6 = 60$ first, then has trouble because $6$ is not prime.

Always start with the smallest prime, $2$ , and only move to the next prime ( $3$ , then $5$ , and so on) when the current prime no longer divides in exactly. This guarantees every divisor you use is prime.

Try one yourself

Your turn

Express $180$ as a product of prime factors using exponent notation. What is the value of the largest exponent in your answer?

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About this worksheet

Why we built it

Powers and prime factors is where Year 7 students move beyond basic multiplication and learn to express numbers in a more powerful, compact form. These worksheets guide students from writing numbers in expanded notation using powers of 10, all the way to breaking any natural number down into a product of prime powers using exponent notation, with worked solutions for every question so they can check their reasoning step by step.

What's covered

Sub-skills your student will practise

✓Writing numbers in expanded notation using powers of 10
✓Understanding base and exponent notation (e.g. 2³ = 2 × 2 × 2)
✓Identifying prime and composite numbers
✓Finding prime factors using factor trees and repeated division
✓Expressing numbers as a product of prime powers (e.g. 360 = 2³ × 3² × 5)
✓Comparing and ordering numbers written in exponent notation

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Frequently asked

Questions parents ask about powers and prime factors

What is prime factorisation and why does Year 7 study it?

Prime factorisation means writing a number as a product of its prime number building blocks, using exponent notation where primes repeat. It underpins later topics like highest common factor, lowest common multiple and simplifying fractions.

What is the difference between a factor tree and repeated division?

Both methods find the same prime factors. A factor tree splits a number into any two factors and keeps branching until all ends are prime. Repeated division divides by the smallest prime each time, working downward until the result is 1. Students should be comfortable with both.

How does expanded notation with powers of 10 connect to place value?

Expanded notation shows each digit multiplied by its place value as a power of 10. For example, 4 506 = 4 × 10³ + 5 × 10² + 0 × 10¹ + 6 × 10⁰. It deepens understanding of why our number system works the way it does.

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