Search simple solutions systematically
Bound a small search, list possibilities in order and test all conditions to find every valid solution.
Section 26-2.2 · 2026–2027 · 2.2 · Find cases satisfying constraints
Original StudyForA scenarios and formative model answers; no Cambridge marks, predicted grades or certified mastery. Alternative reasoned answers may be valid. Full-length unseen papers, a wider mixed-problem bank and calibrated specialist teacher marking remain outstanding.
Before you begin
- Read tables and instructions in English
- Use arithmetic, ratios and whole-number constraints
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Before calculating, what should you identify?
What you will learn
- Enumerate a finite set without missing cases
- Reject cases that violate any given criterion
Turn a guess into a complete search
Define variables, their allowed values and the constraints. Use a table or nested list with one variable changing at a time. Bound each variable from the budget, capacity or allowed digit range. This gives a reason to stop, rather than an arbitrary collection of guesses.
Buy exactly 5 tokens using red tokens at 2 credits and blue tokens at 3 credits, spending exactly 12. Write r+b=5 and 2 r+3 b=12. The possible blue counts are 0 through 5; the costs are 10,11,12,13,14,15, so only b=2 and r=3 work.
Uniqueness needs an exclusion argument
One valid case proves existence, not uniqueness. To claim ‘only’, show all remaining bounded cases fail, or derive an equation whose solution is forced. Avoid double-counting when the order of choices does not matter.
If a proposal violates one requirement, name that requirement. Checking only the total price does not confirm the required item count or any restriction on repeated digits.
Sources for this lesson
Pause and explain
Does finding one valid case prove it is the only case?
Worked example
Find every red/blue purchase in the token problem.
Show the worked solution
- Substitute r=5−b.
- Cost is 2(5−b)+3 b=10+b.
- Set 10+b=12 and check non-negative integers.
Answer The only purchase is 3 red and 2 blue tokens. The equation forces b=2, and this satisfies both the five-token count and the twelve-credit cost.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Guided: Explain why 2 red and 3 blue tokens fail the original problem.
Give me a hint
Check count and cost separately.
Compare my reasoning
- The count is five.
- The price is 2×2+3×3 = 13.
- Compare the result with twelve.
The selection has the correct five-token count but costs 13 credits, one too many. It fails the exact spending constraint.
Look for these in your work
- Checks count
- Checks price
- Names the failed condition
Independent: A two-digit code uses different digits from 1,2,3,4, with digit sum 5. List every code.
Give me a hint
Order matters for a code.
Compare my reasoning
- First digit 1 requires 4; 2 requires 3.
- First digit 3 requires 2; 4 requires 1.
- Verify the digits differ.
The codes are 14, 23, 32 and 41. All four allowed first digits have been considered, so the list is complete; 14 and 41 are different codes.
Look for these in your work
- Lists all four
- Preserves order
- Justifies completeness
Application: Make 14 credits using non-negative numbers of 3-credit and 5-credit vouchers. Find all combinations.
Give me a hint
At most two 5-credit vouchers are affordable.
Compare my reasoning
- For zero 5-credit vouchers, 14 is not divisible by 3.
- For one, 9 remains, giving three 3-credit vouchers.
- For two, 4 remains and is not divisible by 3.
Only three 3-credit vouchers and one 5-credit voucher work: 3×3+5=14. The possible counts of 5-credit vouchers are 0,1,2, and the other two cases fail divisibility.
Look for these in your work
- Bounds the search
- Tests each case
- Verifies the solution
Common mistakes
- Stopping after the first valid case
- Omitting an endpoint
- Counting reordered identical selections twice
What can you explain now?
Does finding one valid case prove it is the only case?
Compare with the explanation
No, the other allowed cases must be excluded
Existence and uniqueness are different claims.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Does finding one valid case prove it is the only case?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.