PRIMARY 4 MATHEMATICS LEARNING GUIDE · BATCH 8 · GUIDE 30
A useful guess is a candidate that can be tested, recorded and improved. Randomly changing numbers until something looks right is different from a systematic search. In a systematic search, we know what can vary, which conditions must stay true, how each trial will be checked and why the remaining possibilities have or have not been covered.
Some Primary 4 problems are solved most clearly by a direct calculation or a bar model. Others become easier when we organise possible values in a list or table. This guide explains how to make that choice, how to prevent double-counting and how to distinguish finding one valid answer from finding every answer.
The MOE Primary Mathematics Syllabus includes strategy development within mathematical problem solving. The small searches, tables and optional challenges below are independent learning activities, not a new compulsory chapter or a set of official examination questions.
Series route: return to the Primary 4 Mathematics Learning Hub. For making a solution readable, see Mathematical Communication and Complete Working.
Navigate: what a search needs · complete lists · tables and informed trials · boundaries and impossible cases · practice laboratory · worked answers · teaching and transfer.
1. Decide whether you are finding one answer or all answers
“Find a rectangle with area 36 square units” asks for one example. A six-by-six square satisfies the condition because a square is also a rectangle. “Find all rectangles with positive whole-number side lengths and area 36, counting rotated copies as the same rectangle” asks for a complete list.
The second task has five answers: one by 36, two by eighteen, three by twelve, four by nine and six by six. The endpoint at six by six marks where the side pairs begin reversing.
A single correct example does not complete an all-answers task. Conversely, listing every possibility may be unnecessary when the question asks for only one valid construction.
Read the command carefully: find one, find all, find the least, find the greatest, or decide whether any exist. These are different mathematical jobs.
Before calculating, write what success would mean. For an all-answers problem, success includes a reason that nothing was missed. For a minimum problem, it includes evidence that smaller candidates fail.
2. Name what may vary and what must remain fixed
Suppose twelve tickets are sold. Each adult ticket costs five dollars and each child ticket costs three dollars. The total collected is 44 dollars. All values are invented for this exercise.
The numbers of adult and child tickets can vary, but together they must always total twelve. Their combined cost must also equal 44 dollars. A trial of four adults and nine children is invalid before we even calculate money because it contains thirteen tickets.
A sensible search chooses an adult count, then determines the child count by subtracting from twelve. This preserves the first condition automatically. The table can then test the second condition.
For other problems, the variable might be a side length, a digit in a position, a number of packets or a departure time. The fixed conditions might be total length, available digits, total items or a repeating interval.
This distinction reduces aimless work. We change a permitted choice while retaining all information the question has already fixed.
3. List in an order that makes omissions visible
Use the digits one, three and five once each to make three-digit numbers. Start with one in the hundreds place and list every arrangement of the remaining two digits: 135 and 153. Then use three: 315 and 351. Finally use five: 513 and 531.
The complete list is 135, 153, 315, 351, 513, 531. We know the list is complete because every allowed hundreds digit has been used, and each case contains both orders of the two remaining digits.
A random list might contain the same six values, but it is harder to inspect for omissions or repetitions. Grouping the cases by the first decision provides a visible reason for completeness.
Do not introduce a formula when a short organised list explains the structure more clearly. The goal at this stage is to control the cases.
When the available choices become too numerous for a practical list, reconsider the representation or look for another mathematical relationship. Systematic listing is a tool, not an instruction to write an enormous table regardless of purpose.
4. Zero changes the rules for the first digit
Use zero, two and five once each to make a three-digit number. Zero cannot occupy the hundreds place because a leading zero would not make an ordinary three-digit number.
With two first, the possibilities are 205 and 250. With five first, they are 502 and 520. Therefore there are four valid numbers.
The strings 025 and 052 use the digits, but they represent 25 and 52 rather than three-digit numbers. They fail a condition of the question.
A good search does not just produce arrangements; it tests which arrangements belong to the requested set. Keep the rule about the first digit visible when listing.
Now change the question to “make three-character codes”. A code might be permitted to begin with zero, depending on the stated rules. The answer can change because the object being formed has changed, even though the same symbols are available.
5. Decide whether order changes the answer
Choosing two colours from red, blue and green without assigning roles gives three selections: red-blue, red-green and blue-green. Blue-red is the same selection as red-blue when order does not matter.
Choosing a colour for the top stripe and a different colour for the bottom stripe gives six designs. Red above blue and blue above red are different because the positions have distinct roles.
The same issue appears in number pairs. For two unnamed numbers that total ten, the pair two and eight may be counted once. For Amir’s amount and Bea’s amount, assigning two to Amir and eight to Bea is a different allocation from the reverse.
State the convention before listing. “Order matters” is not an arbitrary teacher preference; it follows from what the objects represent.
When a student double-counts, ask whether the repeated entries describe genuinely different outcomes or simply different descriptions of the same outcome. That question is more useful than crossing out rows without explaining why.
6. A table can record independent choices
A child chooses one shirt from red, blue and green and one pair of trousers from black and white. Assume all six pairings are allowed.
| Shirt | Black trousers | White trousers |
|---|---|---|
| Red | Red with black | Red with white |
| Blue | Blue with black | Blue with white |
| Green | Green with black | Green with white |
Every row uses one shirt, and every column uses one trouser choice. The six occupied cells represent the six outfits.
The table is complete because each allowed shirt has been paired with each allowed pair of trousers. If a rule excludes green with white, that cell must be removed and the answer becomes five.
Do not multiply choice counts blindly when restrictions make some pairings invalid. The table exposes which combinations actually satisfy the conditions.
This kind of layout is useful when two small choice sets interact. For three or more choices, a branching list or grouped cases may be easier to read.
7. Build a table that preserves a fixed total
Return to the twelve tickets costing five or three dollars, with total collection 44 dollars. Choose the adult count and subtract it from twelve to find the child count.
| Adults | Children | Total collection |
|---|---|---|
| 0 | 12 | $36 |
| 1 | 11 | $38 |
| 2 | 10 | $40 |
| 3 | 9 | $42 |
| 4 | 8 | $44 |
The matching row gives four adult and eight child tickets. Check the conditions independently: four plus eight is twelve, and four lots of five plus eight lots of three total 44.
Why is another later row unnecessary here? Replacing one child ticket with one adult ticket increases the collection by two dollars. After reaching 44, further replacements only increase the total. They cannot return to 44.
This is a justified stopping rule, not simply stopping at the first pleasing result. The direction of change shows why no later candidate can also work.
8. Make guesses that respond to the previous result
Optional challenge. A different event sells twelve tickets at the same prices, but collects fifty dollars. A first trial of four adults and eight children gives 44 dollars, six too little.
Each replacement of a child ticket by an adult ticket adds two dollars without changing the total number of tickets. Three replacements add the missing six. The next informed trial is seven adults and five children.
Check: seven adult tickets cost 35 dollars and five child tickets cost fifteen dollars, giving fifty dollars. The count remains twelve.
A guess-and-check method becomes more efficient when the learner records how a controlled change affects the result. Jumping to unrelated values would discard useful evidence from the first trial.
The adjustment works because the per-ticket difference is fixed. In another problem, increasing a candidate may have a different or irregular effect. Inspect that relationship before choosing the size and direction of the next guess.
9. Sometimes the table reveals a direct method
For the fifty-dollar ticket problem, imagining all twelve tickets as child tickets gives a starting collection of 36 dollars. The required collection is fourteen dollars higher. Each adult replacement adds two dollars, so seven adult tickets are needed.
This direct difference method and the table describe the same structure. The table can help a learner see it before compressing the calculation.
Do not insist that the child keep guessing after the relationship has become clear. Equally, do not demand a shortcut before the learner can explain why each replacement changes the total by two.
A model, table and direct calculation can support one another. Use the representation that makes the changing and fixed quantities easiest to control.
For a comparison of several valid routes, revisit Multiple Solution Routes. Efficiency should come from understanding, not from hiding the conditions.
10. Use factor pairs to organise geometric possibilities
Find all rectangles with positive whole-number side lengths and area 36 square units, counting rotations as the same rectangle. The product of the side lengths must be 36.
Test candidate shorter sides in increasing order. One pairs with 36, two with eighteen, three with twelve, four with nine and six with six. Five does not divide 36 exactly. After the equal pair, the remaining successful pairs would reverse ones already listed.
The five possibilities have perimeters 74, forty, thirty, 26 and 24 units respectively. Thus equal area does not force equal perimeter.
The finite list also answers a minimum question: among these positive whole-number rectangles, the six-by-six square has the smallest perimeter, 24 units.
Notice the scope of that claim. It follows from a complete list of the specified candidates. We are not claiming that listing these five cases proves every possible geometry statement about arbitrary real-number dimensions.
11. Perimeter gives a different search rule
A rectangle has perimeter 26 units and positive whole-number side lengths. Opposite sides are equal, so one length and one width together total thirteen.
Counting rotations once, the possibilities are 1 and 12; 2 and 11; 3 and 10; 4 and 9; 5 and 8; 6 and 7. After six and seven, the next pair reverses a listed pair.
This is not a factor-pair search because the condition fixes a sum rather than a product. The same word “rectangle” appears, but the known quantity selects the relationship.
Calculate a few areas to compare the results: one by twelve gives twelve square units, while six by seven gives 42. Equal perimeter does not force equal area.
Before making a table, identify whether you are preserving a sum, product, difference, fraction or count. A well-organised table built around the wrong relationship remains a wrong solution.
12. Find all packet combinations without overlooking zero
Optional challenge. Small packets contain four counters and large packets contain six. A collection contains exactly 24 counters, with no loose counters. Find all possible packet combinations, allowing either packet type to be absent.
There can be at most four large packets. Test large-packet counts zero through four. The remaining counters must form complete small packets.
| Large packets | Counters remaining for small packets | Valid small-packet count? |
|---|---|---|
| 0 | 24 | 6 |
| 1 | 18 | No whole number |
| 2 | 12 | 3 |
| 3 | 6 | No whole number |
| 4 | 0 | 0 |
The solutions are six small and zero large; three small and two large; zero small and four large. If the question instead required at least one packet of each type, only the middle combination would remain.
Zero is sometimes a valid count and sometimes excluded. Decide from the conditions rather than from an assumption that every named object must appear.
13. Set lower and upper bounds before searching
In the packet problem, negative packet counts are impossible and more than four large packets would exceed 24 counters. Those limits make the search finite.
For twelve tickets, an adult count cannot exceed twelve or be less than zero. For a three-digit number, the first digit cannot be zero. For a rectangle with positive whole-number side lengths and a fixed perimeter, both sides must fit inside the half-perimeter total.
Bounds can also help reject an impossible total. Twelve tickets costing either three or five dollars must collect at least 36 dollars and at most sixty dollars. A claimed total of 65 dollars cannot fit those prices and ticket count.
These checks can be made before calculating every candidate. A boundary argument is often more informative than a long list of failed guesses.
When the question uses “at least” or “at most”, include the boundary if equality is permitted. “Less than twenty” and “at most twenty” do not define exactly the same set.
14. A budget question may have several valid answers
A fictional purchase contains exactly eight items. Notebooks cost three dollars each and pencils cost two dollars each. At least one of each is required, and the budget is at most twenty dollars.
If all eight were pencils, the cost would be sixteen dollars. Replacing a pencil with a notebook adds one dollar. The permitted notebook counts are one, two, three and four.
The valid purchases are one notebook and seven pencils for seventeen dollars; two and six for eighteen; three and five for nineteen; and four and four for twenty.
Stopping at the first valid row would not answer “find all purchases”. Reporting only the twenty-dollar row would silently change “at most twenty” into “exactly twenty”.
The wording determines whether we seek a feasible example, every feasible case or a maximum number of notebooks. A correct table can answer all three questions, but the final sentence must match the one actually asked.
15. Explain why no candidate works
Seven items each cost either two or four dollars. Can their exact total cost be nineteen dollars?
No. Every item contributes an even number of dollars, so the sum of all seven costs is even. Nineteen is odd. The stated prices and count cannot produce that total.
A complete table would also show failure, but the odd-even argument explains every possible combination at once. This is a useful optional reasoning challenge using familiar whole-number properties.
Be careful not to claim impossibility after only two unsuccessful guesses. Failed trials show that those particular candidates fail. They do not eliminate untested candidates unless a justified rule covers them.
For an all-answers task, a proof of completeness may be a bounded table, a branching list or an argument that rules out a whole set. Choose a method whose coverage you can explain.
16. Check the answer and check the search
There are two checks in a search problem. First, test each reported answer against every condition. Second, test whether the search includes every required case without duplicates.
For the twelve tickets and 44 dollars, four adults and eight children satisfy both count and money conditions. The consistent two-dollar increase between rows explains why no later row can also give 44.
For all three-digit numbers formed from zero, two and five, each listed number uses the digits once and is genuinely three-digit. Grouping by the possible non-zero first digits explains completeness.
For rectangles counted without rotation, a pair should not appear again in reverse. The duplicate check is part of the mathematics because the question defines what counts as a distinct answer.
A neat table is not sufficient by itself. Its columns must preserve meaningful quantities, its rows must cover the permitted choices and its stopping point must have a reason.
17. Twenty search and listing tasks
Use a list, table, diagram or direct argument. Explain why the search is complete when the question asks for all possibilities. The ticket, packet and budget tasks are optional challenges for learners ready to coordinate more than one condition.
- Use the digits one, three and five once each to list all three-digit numbers.
- Use zero, two and five once each to list all ordinary three-digit numbers.
- Choose two different colours from red, blue and green, ignoring order. List all choices.
- Choose a top and bottom stripe from those colours, with different colours required. List all designs.
- Three shirt colours and two trouser colours may be paired freely. Show all six outfits in a table.
- List all rectangles of area 36 with positive whole-number side lengths, counting rotations once.
- List all positive whole-number side pairs for a rectangle with perimeter 26, counting rotations once.
- List all common multiples of four and six that are greater than ten and less than forty.
- Twelve tickets cost either five or three dollars each and collect 44 dollars. Find both ticket counts.
- Using the same ticket count and prices, find the counts when the collection is fifty dollars.
- Fifteen vehicles are bicycles with two wheels or tricycles with three wheels. There are 36 wheels altogether. Find each vehicle count.
- Twelve items cost either two or five dollars each. Their total is 39 dollars. Find each count.
- Small packets contain four counters and large packets six. Find every packet combination totalling 24 counters, allowing zero of a type.
- Repeat Question 13 with at least one packet of each type required.
- Exactly eight items are notebooks costing three dollars or pencils costing two. Find all purchases costing at most twenty dollars with at least one of each item.
- Can seven items costing either two or four dollars each total exactly nineteen dollars? Explain.
- Four people A, B, C and D each shake hands with every other person once. List the distinct handshakes.
- List pairs of positive even numbers that total twelve, counting reversed pairs once. Then require the two numbers to be different.
- Two signals flash together at time zero and then every four and six seconds respectively. List their simultaneous flashes after time zero and before thirty seconds.
- A student tests two packet combinations unsuccessfully and declares that no solution exists. What evidence is still needed?
18. Worked answers and completeness checks
1. 135, 153, 315, 351, 513, 531. Each of the three possible first digits has both arrangements of the remaining two digits. This grouped method covers all six cases.
2. 205, 250, 502, 520. The first digit is either two or five. A leading zero would produce a two-digit number rather than an ordinary three-digit number.
3. Red-blue, red-green, blue-green. Reversing the order does not create a new selection because the chosen colours have no different assigned roles.
4. Red-blue, red-green, blue-red, blue-green, green-red, green-blue, with the first colour named at the top. Position makes the reverse design different.
5. Pair each of the three shirts with each of the two trouser choices. There are six occupied cells. The table is complete because every row contains both trouser choices.
6. The side pairs are 1 and 36; 2 and 18; 3 and 12; 4 and 9; 6 and 6. These are the factor pairs of 36, with reverse copies excluded.
7. Half-perimeter is thirteen. The side pairs are 1 and 12; 2 and 11; 3 and 10; 4 and 9; 5 and 8; 6 and 7. Later positive pairs reverse earlier ones.
8. 12, 24, 36. They satisfy both divisibility conditions and the stated open interval. Zero is not included because it is not greater than ten.
9. Four five-dollar tickets and eight three-dollar tickets. The count is twelve and the collection is 20 + 24 = 44 dollars.
10. Seven five-dollar tickets and five three-dollar tickets. Twelve cheaper tickets would collect 36 dollars; seven replacements add fourteen, giving fifty.
11. Nine bicycles and six tricycles. Fifteen bicycles would have thirty wheels. Each replacement by a tricycle adds one wheel, so six replacements produce 36.
12. Seven two-dollar items and five five-dollar items. Twelve cheaper items cost 24 dollars. Each replacement adds three dollars; five replacements add fifteen, giving 39.
13. Six small and zero large; three small and two large; zero small and four large. Testing zero through four large packets is sufficient because five large packets would exceed 24 counters.
14. Only three small packets and two large packets remain valid. The other two solutions fail the newly added requirement that both types must occur.
15. The possibilities are 1 notebook and 7 pencils; 2 and 6; 3 and 5; 4 and 4. Their costs are seventeen, eighteen, nineteen and twenty dollars. More notebooks exceed the budget; zero notebooks violates the condition.
16. No. Every permitted price is even, so any total of those prices is even. An exact total of nineteen dollars is incompatible with the stated prices.
17. AB, AC, AD, BC, BD, CD: six handshakes. BA repeats AB because the handshake is one interaction between the same two people.
18. The pairs are 2 and 10; 4 and 8; 6 and 6. Requiring different numbers removes six and six, leaving the first two pairs.
19. Twelve seconds and 24 seconds. These are the common multiples of four and six strictly after zero and before thirty.
20. The student needs a complete bounded search or a reason covering all untested possibilities. Two failed trials show only that those two combinations do not work.
19. Teach the search plan before increasing the numbers
Begin with a list small enough to inspect in full. Ask the learner to choose the first sorting decision and explain why every allowed choice appears. Then introduce one restriction, such as a forbidden leading zero or an excluded colour pairing.
For a table, ask what each column means before filling rows. A row should preserve the fixed total or other condition by construction whenever possible. This reduces the number of independent guesses.
After a trial misses the target, ask what changed and by how much. Does increasing the candidate move the result in the right direction? Can the next trial use evidence from the previous one?
Delay large searches until the learner can explain completeness in a small one. More candidates are not automatically better practice; they may add copying while leaving the central reasoning unchanged.
A useful final challenge is to ask the child to make a new question for the same table. A table of ticket counts might support finding an exact collection, the greatest affordable adult count or all collections below a limit. The representation remains the same while the requested conclusion changes.
20. Move from solving a search to designing a sound question
When a learner can explain why a list is complete, the next step is to write conditions that create a desired set of answers. Which condition gives one solution? Which leaves several? Which accidentally rules out every candidate?
Continue to Problem Posing, Changing Conditions and Creating Valid Questions. To practise these decisions without an adult supplying the next trial, use Home Learning, Homework Independence and Parent Support.
Final checkpoint: can the learner state the allowed choices, preserve the conditions, organise the cases, justify the stopping point and check every reported result?
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Original small-case searches and optional challenges; school sequence and task instructions take precedence over this suggested practice order.
