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Six variables 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝐹 are distinct integers from 1 to 10 (inclusive).

They satisfy the following conditions (I've added mathematical definitions on the right to avoid ambiguity):

  1. 𝐵 is adjacent to 𝐷 → |B-D| = 1
  2. 𝐶 is not the largest → C < Max(A,B,C,D,E,F)
  3. 𝐴 is closer to 𝐷 than to 𝐶 → |A-D| < |A-C|
  4. No two variables sum to 10 → ∀X,Y∈{A,B,C,D,E,F}(X≠Y⇒X+Y≠10)
  5. 𝐷 is the smallest → D = Min(A,B,C,D,E,F)
  6. 𝐸 < 𝐴

Question: How would you determine the six values step by step, using logic rather than trial and error? i.e. forced deductions that don't begin by assuming the value of any variable.

Note: This puzzle has a special property: it is perfectly entangled. Every one of the six clues is necessary to determine every one of the six variables — remove any single clue and all six variables become ambiguous. No subset of clues pins down even one variable.

Attribution: My unique puzzle, based on the style from my book Six-Figure Logic – Volume II

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2 Answers 2

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We start by noting

The order DBEACF, from the clues 5, 1, 3, 6, and 2.

We then note

A >= 4, |A-D| >= 3 and so C >= 8 by 3.

However,

if C = 8 we have A=4, D=1, B=2, E=3 and so B+C=10 which contradicts 4.

We conclude

C=9, F=10 meaning D cannot be 1 (again by 4). The only possibility left to permit 3 is A=5 D=2. B=3 E=4 immediately follow.

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    $\begingroup$ Wouldn't your However conclusion be against the rules, as it is a forced deduction that begins by assuming the value of a variable? $\endgroup$ Commented 7 hours ago
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Answer:

(A,B,C,D,E,F) = (5,3,9,2,4,10)

Explanation without any “if” statements or case works:

  • Rules 1+5: B = D+1.

  • Rules 3+6: D < B < E < A < C. In particular, A ≥ D+3, C ≥ A+4, and so C ≥ D+7.

  • Rule 2: C < F, so F ≥ D+8, and so D ≤ 2. So either D = 2 or B = 2.

  • Rule 4: none of the numbers is 8. Since C ≥ D+7 ≥ 8, we deduce that C = 9 and F = 10.

  • Rule 4 for (D,C): D = 2, and so B = 3.

  • D+3 ≤ A ≤ C-4 fixes A = 5, so E = 4.

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