PRIMARY 5 MATHEMATICS LEARNING GUIDE · BATCH 13 · GUIDE 51
Logical deduction turns given conditions into consequences that must be true. Elimination removes cases that cannot satisfy those conditions. Together they let a learner solve a problem without guessing every possibility.
This guide develops constraint reading, contradiction checks, impossible cases, parity, divisibility, bounds, mutually exclusive options and proof by exclusion. It is a problem-solving owner rather than a separate syllabus topic.
Return to the Primary 5 Mathematics Learning Hub. Related owners: Systematic Listing, Case Organisation, Tables & Invariant Thinking · Guess, Check, Improve · Mathematical Modelling.
1. A constraint is a condition every valid answer must satisfy
If a two-digit number has digits summing to 11, the pair of digits must satisfy that condition. If the tens digit is also 3 more than the ones digit, only pairs satisfying both conditions remain candidates.
Constraints shrink the search space.
2. Deduction asks what must follow
If a number is even, its last digit must be 0,2,4,6 or 8. If it is also greater than 70 and less than 80, then only 72,74,76,78 remain possible.
Each condition creates a narrower set.
3. Elimination asks what cannot remain
If a candidate violates even one required condition, remove it. There is no need to keep testing a case already known to be impossible.
Efficient reasoning rejects early.
4. Contradiction is a powerful signal
Suppose a trial leads to “the remainder is 37 when dividing by 34”. That cannot be a valid remainder because another complete group could be formed. The contradiction proves the trial or earlier reasoning is wrong.
5. Proof by exclusion
If exactly four cases are possible and three are eliminated by valid reasons, the remaining case must be the answer.
The learner should record why each rejected case fails, not merely cross it out.
6. Use bounds first
If 15 objects each cost either $6 or $10, the total must be between $90 and $150. A claimed total of $180 is impossible without further work.
Bounds can eliminate entire problem states immediately.
7. Use parity
If a total starts even and every move changes it by 2, it remains even. Therefore an odd target is impossible.
Parity can replace long simulation with one invariant statement.
8. Use divisibility
If objects must be split into equal groups of 6 with none left, the total must be divisible by 6. A total of 74 cannot satisfy that exact grouping condition.
9. Use whole-object constraints
A solution requiring 3.5 buses is not operationally valid if buses are indivisible. A solution requiring 12.4 students is impossible because people are whole objects.
10. Use geometry constraints
Triangle angles must total 180°. A proposed triangle with angles 90°,60°,50° is impossible because the total is 200°.
Geometry facts eliminate cases before measurement or algebra is attempted.
11. Use capacity constraints
A 30-litre tank cannot contain 42 litres without overflow. A model predicting 42 litres inside the tank violates capacity unless the problem allows overflow.
12. Use order constraints
If A must be before B and B before C, then C cannot be before A. Simple ordering rules can eliminate many arrangements in scheduling and queue problems.
13. Use mutually exclusive categories
If every object is exactly one of red or blue, no object can be counted in both groups. If totals overlap, the model violates the category rule.
14. Use exhaustive categories
If red, blue and green are the only possible colours, their counts must add to the total. If they add to less, a category is missing; if more, there is overlap or error.
15. Use minimum and maximum values
If five positive whole numbers must total 12, the minimum contribution of the other four numbers can bound any chosen one. Such bounds restrict feasible cases.
16. Use difference constraints
If A and B differ by 20, any proposed pair must preserve that gap. The pair 80 and 50 can be rejected immediately because the difference is 30.
17. Use total constraints
If A+B=150, any candidate pair not summing to 150 is impossible regardless of any other attractive pattern.
18. Combine constraints progressively
Example: Find a two-digit even number greater than 50, digits sum to 9, and tens digit exceeds ones digit.
Even candidates above 50 include many values. Digit-sum 9 reduces them to 54,72,90. Tens digit greater than ones eliminates none of these except 90 still qualifies. If another condition says number less than 80, 90 is eliminated. Candidates 54 and72 remain.
Each condition should visibly shrink the set.
19. Logical deduction can reveal uniqueness
Suppose a number is between 60 and 70, even, and digit sum is 10. Candidates 62,64,66,68. Digit sums are 8,10,12,14. Only 64 works. The conditions prove uniqueness.
20. Contradiction can test an assumption
Assume a quantity is larger than another. If this forces a negative number of items later, the assumption may be impossible. Try the alternative relation.
This is controlled reasoning, not arbitrary switching.
21. Elimination in multiple-choice questions
Options can often be rejected using scale, units, parity, sign or bounds before exact calculation. If a percentage below 100% of 480 is requested, any option above 480 is impossible.
Elimination should support—not replace—mathematical reasoning.
22. Eliminate by units
If the target is area, an option labelled cm cannot be correct; square units are required. If the target is rate, a pure number without the required per-unit meaning may be incomplete.
23. Eliminate by magnitude
If 49% of 398 is near 200, options 19.5 and 1950 can be rejected quickly. Estimation turns into logical elimination.
24. Eliminate by inverse check
If x is claimed to solve 5x−18=102, substitute it. A candidate failing the original condition is impossible.
25. Deduction tables
| Candidate | Constraint 1 | Constraint 2 | Keep? |
|---|---|---|---|
| 54 | Even ✓ | Digit sum 9 ✓ | Yes |
| 63 | Even ✗ | Digit sum 9 ✓ | No |
| 72 | Even ✓ | Digit sum 9 ✓ | Yes |
Tables help ensure every candidate is judged by the same rules.
26. Deduction trees
For branching cases, draw one branch per decision and prune branches as soon as they violate a constraint. This prevents wasting time completing impossible paths.
27. Impossible cases are informative
Rejecting a case teaches us which condition controls the problem. A contradiction is not wasted work if it narrows the remaining possibilities.
28. Distinguish “not yet proven” from “impossible”
A case should be eliminated only when it violates a condition or leads to contradiction. “I do not see how it works” is not evidence of impossibility.
Logical discipline matters.
29. Distinguish necessary and sufficient conditions
Being even is necessary for divisibility by 6 but not sufficient: 8 is even but not divisible by 6. A learner should not stop after checking only one necessary condition.
30. Use counterexamples against false rules
If someone claims “every even number is divisible by 4”, the counterexample 6 disproves the claim. One valid counterexample is enough to reject a universal statement.
This connects deduction with the Special Cases and Counterexamples guide.
31. Error map
| Error | Cause | Repair question |
|---|---|---|
| Crosses out case without reason | Elimination not justified | Which condition does this case violate? |
| Stops after checking one constraint | Necessary condition mistaken for sufficient | Does it satisfy every condition? |
| Calls difficult case impossible | No contradiction established | What exact condition fails? |
| Lists too many cases | No early pruning | Which bound, parity or divisibility rule can eliminate branches sooner? |
| Accepts final remaining case without checking | Earlier elimination may contain error | Does the survivor satisfy all original conditions? |
32. Deduction routine
- List the conditions.
- Convert each condition into a mathematical constraint.
- Apply the strongest easy constraints first.
- Eliminate impossible cases with reasons.
- Continue until one or a small number remain.
- Verify the survivor against every original condition.
33. Practice laboratory A
- Find the number between 60 and70 that is even and has digit sum 10.
- Can 74 objects be split into equal groups of 6 with none left?
- Can a triangle have angles 90°,60°,50°?
- 15 items cost either $6 or $10. Can total cost be $170?
- A 30-L tank is predicted to hold 42 L without overflow. What constraint fails?
34. Practice laboratory B
- A two-digit number is even, greater than 70, less than 80, digit sum 13. Find it.
- A+B=150 and A−B=20. Which candidate pair is possible: (85,65), (90,60), (100,50)?
- A number is divisible by 6. Which must be true: even, divisible by3, both, or neither?
- Four arrangements remain after initial filtering. Three violate a seating constraint. What must be done before accepting the fourth?
- Disprove: “Every multiple of 3 is odd.”
35. Answers
1. 64.
2. No; 74 is not divisible by6.
3. No; angles total 200°.
4. Maximum=15×10=$150, so impossible.
5. Capacity bound.
6. Candidates 72,74,76,78; digit sums 9,11,13,15 → 76.
7. (85,65) sums 150 and differs by20 → (85,65).
8. Both.
9. Verify the remaining case satisfies all original constraints and that eliminations were valid.
10. Counterexample: 6 is a multiple of3 and is even.
36. Full deduction problem
A class buys exactly 20 tickets. Each ticket costs either $8 or $12. Total spending is $208. Without starting from an equation, use logical constraints and elimination to determine the number of $12 tickets.
Let p be premium tickets. Since all 20 tickets at $8 would cost $160, spending is $48 above that minimum. Replacing one $8 ticket with one $12 ticket raises total by $4. Therefore the number of replacements must satisfy 4p=48, giving p=12.
Constraint interpretation:
- p must be a whole number from 0 to20;
- total cost changes only in multiples of $4 above $160;
- $208−$160=$48 is exactly 12 replacement steps.
Every other premium-ticket count produces a different total, so the solution is unique.
37. Final checkpoint
A strong Primary 5 learner can translate conditions into constraints, use bounds, parity, divisibility, geometry and capacity to reject impossible cases, distinguish justified elimination from mere difficulty, prune search trees early, use contradiction and counterexamples carefully, and verify the final surviving case against every original condition.
Continue to Primary 5 Mathematics Learning Guide | Special Cases, Boundary Cases, Counterexamples & Testing General Claims.
Wintour House V1.0 · CivDJ · eduKate Publishing: every elimination needs a reason, every survivor needs verification, and the shortest proof is often the constraint that makes all other cases impossible.
