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H3 problems often have no routine method. This lesson collects strategies that help when you are stuck, starting with working backwards from the result you want. It then separates a conjecture, suggested by a pattern in examples, from a proof, and practises the skill the 9820 paper asks for directly: reading someone else’s solution, deciding whether each step is justified, and repairing it where it fails.
Work backwards from the end
When a problem has a clear finishing condition, it often helps to start there and work backwards. Other heuristics in this lesson: solve a simpler problem, look for pattern and structure, consider cases, and restate the problem, for example by its contrapositive.
Worked example 1
A take-away game
Worked example 1
A take-away game
There are 21 counters. Two players take turns to remove 1, 2 or 3 counters, and whoever takes the last counter wins. Should you go first, and if so, how should you play?
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Work backwards from the end
Method
Call a number of counters losing if the player about to move loses against best play.Reason
0 counters is losing for the player to move, because the opponent has just taken the last one.Working
Answer
From 1, 2 or 3 you can move to 0, so they are winning. From 4, every move leaves 1, 2 or 3, so 4 is losing. Repeating, the losing positions are 0,4,8,12,16,20: the multiples of 4.
Turn the pattern into a strategy
Method
From a non-multiple of 4, take enough to reach a multiple of 4.Reason
From a multiple of 4, any move of k counters can be answered with 4-k, returning to a multiple of 4.Working
Answer
21 is not a multiple of 4, so go first and take 1, leaving 20. After each opponent move of k, take 4-k. You take the last counter.
Quick check
The pattern from small cases became a proof only with the answering move 4-k, which shows the multiples of 4 stay losing for ever. Patterns alone are conjectures.
Check your understanding 1
Extend and generalise
Check your understanding 1
Extend and generalise
Now each player may remove between 1 and k counters. Conjecture which starting numbers are losing for the first player, and justify your conjecture.
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The losing positions are the multiples of k + 1. From a multiple of k + 1, any move removes between 1 and k and leaves a non-multiple; the opponent then removes the remainder on division by k + 1, which is between 1 and k, to return to a multiple. From a non-multiple, removing that remainder reaches a multiple. The special case k = 3 gives the multiples of 4.
A pattern is a conjecture, not a proof
Place n points on a circle, join every pair with a chord, and choose the points so that no three chords meet at one interior point. Count the regions inside the circle:
| n | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Regions | 1 | 2 | 4 | 8 | 16 | 31 |
The first five values suggest 2ⁿ⁻¹, but n = 6 gives 31, not 32. The correct count is (n choose 4) + (n choose 2) + 1: each crossing of two chords, fixed by 4 of the points, adds a region, as does each chord, fixed by 2 points. Checking the formula: for n = 6, 15 + 15 + 1 = 31.
A good investigation formulates a conjecture from cases, tests it on further cases, especially special or extreme ones, and then either proves it or finds the case that breaks it.
Reading and critiquing a solution
To critique a solution, check each line: does it follow from the previous ones, and is every step that changes an equation reversible? A step that is not reversible, such as squaring both sides, can introduce answers that do not satisfy the original.
Spot the mistake 1
An extra solution
Spot the mistake 1
An extra solution
A student solves square root of (x + 2) = x as follows: “Squaring, x + 2 = x², so x²-x-2 = 0, (x-2)(x + 1) = 0, and the solutions are x = 2 and x = -1.” Critique the solution and complete it correctly.
Write your answer, then compare it with the solution
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Find the irreversible step
Method
Squaring is not reversible: a² = b² also allows a = -b.Reason
Candidates from the squared equation must be checked in the original.Working
Answer
x = 2: square root of 4 = 2, valid. x = -1: square root of 1 = 1 ≠ -1, so it is not a solution; it solves - square root of (x + 2) = x instead. The only solution is x = 2.
Common mistake
The algebra after squaring is correct; the error is in the logic of the claim “the solutions are”. Since square root of (x + 2) ≥ 0, the condition x ≥ 0 could also have been stated at the start, ruling out x = -1 immediately.
Common mistakes
- Presenting a pattern checked on several cases as a proof.
- Working backwards from the conclusion and then writing the steps in that order, so that the argument assumes what it should prove.
- Accepting a step in a critiqued solution because its conclusion is true.
- Stopping at “the solution is wrong” without identifying the first invalid step and correcting it.
- Testing a conjecture only on convenient cases, such as small positive integers, and missing a counterexample.
Before you move on
Exam-style question 1
Sums of consecutive positive integers
Exam-style question 1
Sums of consecutive positive integers
Some positive integers can be written as a sum of two or more consecutive positive integers; for example, 9 = 4 + 5 = 2 + 3 + 4.
(a) By investigating the integers from 1 to 16, form a conjecture about which positive integers cannot be written in this way. [2]
(b) Prove your conjecture. [5]
Write your answer, then compare it with the solution
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Investigate and conjecture
Method
Try each integer: 3 = 1 + 2, 5 = 2 + 3, 6 = 1 + 2 + 3, 7 = 3 + 4, 9 = 4 + 5, 10 = 1 + 2 + 3 + 4, 11 = 5 + 6, 12 = 3 + 4 + 5, and so on.Reason
Listing small cases systematically shows which values never appear.Working
Answer
From 1 to 16 the exceptions are 1,2,4,8,16. Conjecture: a positive integer cannot be written as a sum of two or more consecutive positive integers exactly when it is a power of 2.
Find a formula for such a sum
Method
The sum of k ≥ 2 consecutive integers starting at a ≥ 1 is N = (k(2a + k-1))/2.Reason
A formula turns “can be written” into a statement about factors of 2N.Working
Answer
So 2N = k(2a + k-1). The two factors differ by 2a-1, which is odd, so one is even and the other odd; and both are at least 2, since k ≥ 2 and 2a + k-1 ≥ k + 1.
Powers of 2 cannot be written this way
Method
If N = 2^j, then 2N = 2^(j + 1) has no odd factor greater than 1.Reason
But the odd factor in k(2a + k-1) is at least 3.Working
Answer
This is a contradiction, so no power of 2 is such a sum.
Every other integer can
Method
If N = 2^jm with m ≥ 3 odd, let p be the smaller and q the larger of 2^(j + 1) and m.Reason
Choosing k = p and 2a + k-1 = q reverses the formula; the opposite parity of p and q makes a a whole number.Working
Answer
Then pq = 2N, p ≥ 2, and a = (q-p + 1)/2 is a positive integer because q-p is odd and positive. For example, N = 12 gives 2N = 24 = 8 × 3, so k = 3, a = 3: 3 + 4 + 5 = 12. blacksquare
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Tick a point only if your written work contains it. This self-check does not record a mark.
Common mistake
Checking up to 16 supports the conjecture but proves nothing; the chord-regions pattern above failed at the sixth case. A complete proof needs both directions: that powers of 2 never work, and that every other integer does.
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