This set practises the five skills in International Mathematics investigations: generating cases, forming a conjecture, finding a counterexample, explaining with algebra and stating limits. The questions go from easy to harder. Write full working and a sentence of conclusion for each one before opening its answer.
If a question involves repeated percentage change, the percentage-base explorer can help you generate cases. Use the non-calculator working trainer to double-check arithmetic afterwards.
Part A: Generating cases
Q1. Six teams play a tournament where every team plays every other team once. List the matches systematically and find the total.
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Name the teams A to F. A plays 5 others, B plays 4 more, C plays 3 more, D plays 2 more, E plays 1 more.
Total 5 + 4 + 3 + 2 + 1 = 15 matches. Check: 6 × 5 / 2 = 15.
Q2. A row of n squares is made from matchsticks, with 4 sticks for one square and 7 for two squares. Find the number of sticks for 8 squares by building a table, then check with a second method.
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Each extra square adds 3 sticks, so the counts are 4, 7, 10, 13, 16, 19, 22, 25 for n = 1 to 8.
Second method: 8 squares have 9 vertical sticks and 16 horizontal sticks, and 9 + 16 = 25. The answer is 25 sticks.
Q3. A pattern has terms 7, 11, 15, 19 for n = 1 to 4. Find a rule and the term for n = 25.
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Differences are a constant 4, so the rule starts 4n. At n = 1 we need 7, so add 3: rule 4n + 3.
Check: n = 4 gives 19. For n = 25: 4 × 25 + 3 = 103.
Part B: Forming conjectures
Q4. The sum of the first n odd numbers for n = 1 to 5 is 1, 4, 9, 16, 25. State a conjecture and predict the sum for n = 12.
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The values are square numbers, so the conjecture is that the sum is n².
For n = 12 the sum is 144. Check by pairing: the 12 odd numbers 1 to 23 make 6 pairs each summing to 24, and 6 × 24 = 144.
Q5. A table gives 3, 8, 15, 24 for n = 1 to 4. Form a conjecture for the nth term and predict n = 5.
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Differences are 5, 7, 9, which increase by 2, so the rule is quadratic. Values match n(n + 2): 1 × 3, 2 × 4, 3 × 5, 4 × 6.
For n = 5: 5 × 7 = 35. Check with differences: next difference 11, and 24 + 11 = 35.
Part C: Counterexamples
Q6. Conjecture: if n is an odd prime number, then n + 2 is also prime. Find a counterexample.
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n = 3 gives 5 and n = 5 gives 7, both prime. n = 7 gives 9 = 3 × 3.
n = 7 is a counterexample.
Q7. Conjecture: for all positive numbers a and b, √(a + b) = √a + √b. Find a counterexample with whole numbers.
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Take a = 9 and b = 16. Then √(9 + 16) = √25 = 5, but √9 + √16 = 3 + 4 = 7.
a = 9, b = 16 is a counterexample, because 5 ≠ 7.
Q8. Conjecture: if n is prime, then 2n − 1 is prime. The values for n = 2, 3, 5, 7 are 3, 7, 31, 127, all prime. Show that n = 11 is a counterexample. (Hint: try dividing by 23.)
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211 = 2048, so 211 − 1 = 2047. Now 23 × 89 = 23 × 90 − 23 = 2070 − 23 = 2047.
So 2047 = 23 × 89 is not prime, and n = 11 is a counterexample.
Part D: Explaining with algebra
Q9. Show that the sum of any three consecutive even numbers is a multiple of 6.
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Let the numbers be 2n, 2n + 2, 2n + 4. The sum is 2n + 2n + 2 + 2n + 4 = 6n + 6 = 6(n + 1).
This is 6 times a whole number, so it is a multiple of 6. Check: 10 + 12 + 14 = 36 = 6 × 6.
Q10. A pattern has terms 5, 8, 11, 14, which follow 3n + 2. Show that 101 is in the pattern and that 100 is not.
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Set 3n + 2 = 101. Then 3n = 99 and n = 33, a whole number, so 101 is the 33rd term.
Set 3n + 2 = 100. Then 3n = 98 and n = 32.67, not a whole number, so 100 is not in the pattern.
Part E: Stating limits
Q11. A student checks n² + n + 11 for n = 1 to 9 and gets 13, 17, 23, 31, 41, 53, 67, 83, 101, all prime. The student writes “so it is always prime”. Find a counterexample and rewrite the conclusion.
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Try n = 10: 100 + 10 + 11 = 121 = 11 × 11, which is not prime.
n = 10 is a counterexample. A fair conclusion: “n² + n + 11 gives a prime for n = 1 to 9, but not for n = 10, so the rule is not always prime.”
If you got these wrong
Match the kind of error to the lesson, then retry a similar question a few days later.
| What went wrong | Questions | Go to |
|---|---|---|
| Counts were messy or skipped cases | Q1, Q2, Q3 | Generate systematic cases |
| Rule did not fit or was not tested on a new case | Q4, Q5 | Move from observed cases to a conjecture |
| Stopped after agreeing cases, or chose a test value that did not meet the conditions | Q6, Q7, Q8 | Test a counterexample |
| Wrote the algebraic form wrongly, or left the conclusion unstated | Q9, Q10 | Explain a general rule with algebra |
| Claimed “always” without proof | Q11 | State limits of a conclusion |
Record each slip in the mistake log and retest queue so you can return to the same error type later. Back to the module overview if you want the study order again.
If the same type of error keeps returning, it usually points to one habit, not five separate problems. A teacher can look for that habit during online one-to-one Mathematics tuition.