Factor/divisibility algorithms | Year 5

01 / LEARN

Understand the idea.

An algorithm needs precise steps

An algorithm is an ordered set of instructions that another person can follow without guessing. To find factor pairs, start with candidate $1$ , divide the target by it, record a pair if the division is exact, then increase the candidate by $1$ .

Before each test, compare the candidate multiplied by itself with the target. Stop only when that square is greater. Beyond this point the candidate is the larger partner, so its smaller partner would already have been tested.

Tracing the factor-pair algorithm for 20
Candidate	Division	Record a pair?
1	20 ÷ 1 = 20	Yes: 1 and 20
2	20 ÷ 2 = 10	Yes: 2 and 10
3	20 ÷ 3 has a remainder	No
4	20 ÷ 4 = 5	Yes: 4 and 5

Test the rule on different numbers

A square number needs special care: for $16$ , candidate $4$ must be tested because $4 \times 4 = 16$ . Record the pair $4$ and $4$ once. The stopping condition is “greater than”, not “greater than or equal to”.

A procedure with decisions must explain every branch. To test $6$ , reject odd numbers first. For an even number, test whether its digit sum is divisible by $3$ ; accept only if it passes. A failed condition must lead to a clear answer.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Trace a factor search

Use the procedure to list all factor pairs of 30.

Worked solution

1
Set the range of candidates.
$5 \times 5 = 25 \leq 30, 6 \times 6 = 36 > 30$
Test 1 through 5; stop before testing 6.
2
Test candidates one, two and three.
$1 \times 30 = 30, 2 \times 15 = 30, 3 \times 10 = 30$
Each exact division gives one factor pair.
3
Test candidates four and five.
$30 = 4 \times 7 + 2, 5 \times 6 = 30$
4 leaves a remainder, but 5 has an exact partner.
4
Record all successful pairs once.
$1 \times 30, 2 \times 15, 3 \times 10, 5 \times 6$
The stopping rule ensures any later pair would repeat an earlier one.

Answer

1 × 30, 2 × 15, 3 × 10 and 5 × 6.

EXAMPLE 02

Find a missing stopping instruction

A rule says, 'Test one, two, three and so on. Record factors.' What instruction should be added so the rule stops without missing any factor pairs?

Worked solution

1
Identify when a factor pair starts repeating.
Once the candidate exceeds its partner, the smaller partner has already been tested.
2
Write the condition to check before each test.
$candidate \times candidate > target$
This condition means the candidate is beyond every possible smaller partner.
3
Keep an equal-square case.
$5 \times 5 = 25$
For target 25, candidate 5 must still be tested; stop only when the square is greater.

Answer

Stop when candidate × candidate is greater than the target number.

03 / YOUR TURN

Try it yourself.

Check your understanding

For a target of 18, which candidates must be tested before the square stopping rule applies?

Show answer & reasoning

1, 2, 3 and 4.

Test $1$ , $2$ , $3$ and $4$ : $4 \times 4 = 16 \leq 18$ . The next candidate has $5 \times 5 = 25 > 18$ , so the procedure stops before testing $5$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Use a factor-pair procedure to find all factor pairs of 16.
2 marks
For a target of 28, what is the largest candidate tested using the square stopping rule?
2 marks
A learner records $1 \times 24$ , $2 \times 12$ and $4 \times 6$ , but misses one pair. Which candidate was skipped?
2 marks
Follow the two-decision test for six on the number 123. State where the procedure stops.
2 marks
Test the procedure on 25. Explain why $5 \times 5$ appears once rather than twice.
2 marks
Design a short decision rule for divisibility by 10, then use it on 240 and 245.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Use a factor-pair procedure to find all factor pairs of 16.
2 marks
Test one through four. Include the equal-factor pair.
$1 \times 16$ , $2 \times 8$ and $4 \times 4$ .
For a target of 28, what is the largest candidate tested using the square stopping rule?
2 marks
$5 \times 5 = 25$ is not greater than twenty-eight.
$6 \times 6 = 36$ is greater, so stop before six.
5.
A learner records $1 \times 24$ , $2 \times 12$ and $4 \times 6$ , but misses one pair. Which candidate was skipped?
2 marks
$3 \times 8 = 24$ , so the missing pair is three and eight.
3.
Follow the two-decision test for six on the number 123. State where the procedure stops.
2 marks
123 is odd, so there is no need for the second check.
It stops at the even-number test: 123 is not divisible by six.
Test the procedure on 25. Explain why $5 \times 5$ appears once rather than twice.
2 marks
Test one through five.
Five multiplied by itself gives twenty-five; it is a valid pair.
The pairs are $1 \times 25$ and $5 \times 5$ . The equal pair has no different reversed arrangement.
Design a short decision rule for divisibility by 10, then use it on 240 and 245.
3 marks
The final digit is enough for this divisor.
State both possible outcomes so the rule is complete.
Check the final digit. If it is zero, divisible by ten; otherwise not. 240 passes and 245 fails.

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Curriculum reference & lesson scope

Australian Curriculum v9.0; SA Curriculum Mathematics 2026

Factor pairs; multiples; divisibility; factor/divisibility algorithms.

Official curriculum source ↗

In this lesson

Understand the idea.

An algorithm needs precise steps

Test the rule on different numbers

See the method in action.

Trace a factor search

Set the range of candidates.

Test candidates one, two and three.

Test candidates four and five.

Record all successful pairs once.

Find a missing stopping instruction

Identify when a factor pair starts repeating.

Write the condition to check before each test.

Keep an equal-square case.

Try it yourself.

Check your understanding

Keep practising