The idea, step by step
An algorithm needs precise steps
An algorithm is a clear procedure someone else can follow. To find factor pairs of twenty, start with candidate one. Divide twenty by the candidate. If there is no remainder, record the candidate and its partner. Increase the candidate by one and repeat.
A useful stopping point is when the candidate multiplied by itself becomes larger than the target. Beyond that point, any successful pair would repeat one already considered. For twenty, test one through four; stop before five because 5 × 5 is greater than twenty.
| 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 good procedure works for square numbers too. For sixteen, include candidate four because 4 × 4 equals sixteen. Record four and four as one pair, not two new pairs. Try a prime number such as thirteen: only one and thirteen form a pair.
A divisibility procedure can also use decisions. To test six, first ask whether the number is even. If not, stop with 'not divisible by six'. If it is even, test the digit sum for three. State what happens in every branch so the instructions never become ambiguous.
A common trap
Stopping before testing the square-root pair of a square number.
Instead: Stop only when the square is greater than the target. For twenty-five, test five and include 5 × 5.
See the thinking
Trace a factor search
Use the procedure to list all factor pairs of 30.
- Test candidates one through five, because 6 × 6 is greater than thirty.
- One, two, three and five divide thirty without a remainder; four does not.
- Record each matching partner once.
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?
- An endless list of candidates never finishes.
- Pairs repeat once the small factor passes the large factor.
- Stop before testing a candidate whose square exceeds the target; keep an equal square case.
Check your understanding
For a target of 18, which candidates must be tested before the square stopping rule applies?
Show answer
1, 2, 3 and 4.
Four squared is sixteen, which is still below eighteen. Five squared is twenty-five, so stop before five.
Your turn
6 questions · 13 marks · Work at your own pace.
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 × 24, 2 × 12 and 4 × 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 × 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 × 16, 2 × 8 and 4 × 4. - For a target of 28, what is the largest candidate tested using the square stopping rule?2 marks
5 × 5 = 25 is not greater than twenty-eight.
6 × 6 = 36 is greater, so stop before six.
5. - A learner records 1 × 24, 2 × 12 and 4 × 6, but misses one pair. Which candidate was skipped?2 marks
3 × 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 × 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 × 25 and 5 × 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.
Curriculum reference & lesson scope
Australian Curriculum v9.0; SA Curriculum Mathematics 2026
Factor pairs; multiples; divisibility; factor/divisibility algorithms.
Official curriculum source ↗