Six Claim Files and a Single Decisive Clue
Six claim files - A, B, C, D, E and F - are stacked one per shelf on a rack whose shelves are numbered 1 (lowest) to 6 (highest). The sub-editor's note records four conditions: (i) A is kept immediately above D; (ii) exactly two shelves lie between B and E; (iii) C is on an even-numbered shelf but not on shelf 6; (iv) F is on a shelf above C. Before the audit, the team must record the exact stack, not a set of possibilities. A clerk proposes four additional conditions, and only one of them may be adopted: (P) E is on shelf 1; (Q) B is immediately above C; (R) C is on shelf 2; (S) F is immediately below A. The team also needs a count of the arrangements that satisfy conditions (i) to (iv) as they stand, because a clue that leaves two or more live arrangements is useless for audit purposes, while a clue that contradicts the original four is worse than useless.
1. How many arrangements satisfy conditions (i) to (iv), and what are they?
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Step 1 - what condition (iii) permits. C is even and not 6, so C is on shelf 2 or shelf 4.
Step 2 - branch on C.
C = 2. A must be immediately above D and neither may use shelf 2, so (A,D) is (4,3), (5,4) or (6,5).
- (4,3): shelves 2, 3, 4 are occupied; among the free shelves 1, 5, 6 no two are three apart, so B and E cannot be placed. Reject.
- (5,4): free shelves 1, 3, 6. B/E must be (3,6) or (6,3); the leftover shelf 1 then holds F, which is below C, breaching (iv). Reject.
- (6,5): free shelves 1, 3, 4. B/E is (1,4) or (4,1), and F = 3, which is above C = 2. Accept - two solutions.
C = 4. (A,D) is (2,1), (3,2) or (6,5).
- (2,1): free shelves 3, 5, 6. B/E is (3,6) or (6,3); F = 5, above 4. Accept - two solutions.
- (3,2): free shelves 1, 5, 6; no pair three apart. Reject.
- (6,5): free shelves 1, 2, 3; no pair three apart. Reject.
Step 3 - the four live arrangements.
- B1 C2 F3 E4 D5 A6
- E1 C2 F3 B4 D5 A6
- D1 A2 E3 C4 F5 B6
- D1 A2 B3 C4 F5 E6
So the four conditions leave four arrangements, not one.
2. Which of P, Q, R and S should the team adopt, and why do the other three fail?
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Each candidate is tested against the four live arrangements; the right clue must leave exactly one.
- (P) E on shelf 1: only arrangement 2 survives. Arrangement 1 puts B on 1, arrangement 3 puts D on 1, arrangement 4 puts D on 1. Exactly one arrangement - adopt P.
- (Q) B immediately above C: in arrangements 1–2, C is on 2 and B is on 1 or 4; in arrangements 3–4, C is on 4 and B is on 6 or 3. B is never one shelf above C, so Q contradicts conditions (i)–(iv) and leaves zero arrangements.
- (R) C on shelf 2: leaves arrangements 1 and 2, i.e. two arrangements - the ambiguity survives, so R does not serve the audit.
- (S) F immediately below A: in arrangements 1–2, A is on 6 and F on 3; in arrangements 3–4, A is on 2 and F on 5. Never adjacent, so S also gives zero arrangements.
Conclusion: adopt (P). The test of a useful extra clue is not that it sounds specific but that it intersects the live solution set in exactly one point; Q and S are traps because they are precise but unsatisfiable, and R is precise but redundant.
3. Ignoring any extra clue, which one of the following must be true in every arrangement: (a) D is on an odd-numbered shelf; (b) A and C are on consecutive shelves; (c) B is above E; (d) F is on shelf 3?
Read the worked answer
Check each option against all four live arrangements.
| Option | Arr 1 | Arr 2 | Arr 3 | Arr 4 | Verdict |
|---|---|---|---|---|---|
| (a) D odd | D5 ✓ | D5 ✓ | D1 ✓ | D1 ✓ | always true |
| (b) A, C consecutive | 6,2 ✗ | 6,2 ✗ | 2,4 ✗ | 2,4 ✗ | never true |
| (c) B above E | 1>4 ✗ | 4>1 ✓ | 6>3 ✓ | 3>6 ✗ | sometimes |
| (d) F on 3 | ✓ | ✓ | F5 ✗ | F5 ✗ | sometimes |
Answer: (a). The reasoning is structural: C must sit on an even shelf, and A/D immediately above one another can only fit on (6,5) when C = 2 and on (2,1) when C = 4 - in both cases D lands on 5 or 1. Options (b), (c) and (d) are each true in some but not all arrangements, which is exactly the trap the puzzle sets for a candidate who solves one branch and stops.