Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁴ᵗʰa ³ʳᵈb ¹ˢᵗc ⟩, a > b > c
5th → a, 2nd → b, ab=2+4n
⟨ ▧ ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≤ Σ▢
⛔Avoid
⟨ ▧ ▧ − ▢ − − ⟩, Σ▧ ≤ ▢
⟨⋯ Perm(1,3,4) ⋯⟩
#125034_v2.9
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│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ 2 │ │ │ │ │ │▒
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Step 2 │ 2 │ │ 1 │ │ │ │▒
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Step 3 │ 2 │ │ 1 │ │ 0 │ │▒
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Step 4 │ 2 │ │ 1 │ 5 │ 0 │ │▒
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Step 5 │ 2 │ 4 │ 1 │ 5 │ 0 │ │▒
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Step 6 │ 2 │ 4 │ 1 │ 5 │ 0 │ 3 │▒
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Proof of 2025-07-22 WR
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Notation: if nth -> a, then we write [nth] = a.
We consider where 0 is placed. To match ✅「⟨ ⁴ᵗʰa ³ʳᵈb ¹ˢᵗc ⟩, a > b > c」, it cannot be at 4th or 3rd; and to match ✅「5th → a, 2nd → b, ab=2+4n」, it cannot be at 5th or 2nd. Therefore,
(1) 0 = [1st] or [0th].
From this, we consider which two nonzero digits the set {[2nd], [1st], [0th]} should have in order to match ✅「⟨ ▧ ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≤ Σ▢」.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
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│ │ │ │ ▬ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
After trying the possible combinations like {5,4}, {5,3}, etc., we find that only two combinations work:
(2) {[2nd], [1st], [0th]} = {5,4,0} or {5,3,0}.
A fortiori,
(3) 2 ∉ {[2nd], [1st], [0th]}.
Having this, we consider how to match the pattern ✅「5th → a, 2nd → b, ab=2+4n」:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ? │ │ │ ? │ │ │
└───┴───┴───┴───┴───┴───┘
To match it, we need
(4) {[5th], [2nd]} = {2,1} or {2,3} or {2,5}.
Accordingly, we need 2 ∈ {[5th], [2nd]}. Combining this with (3), we get 2 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, we consider where 1 is placed. By (2), it is at 4th or 3rd. But if 1 = [4th] then we cannot match ✅「⟨ ⁴ᵗʰa ³ʳᵈb ¹ˢᵗc ⟩, a > b > c」. Therefore, 1 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │▒
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Step 2 │ 2 │ │ 1 │ │ │ │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Plainly, ✅「⟨ ⁴ᵗʰa ³ʳᵈb ¹ˢᵗc ⟩, a > b > c」 then gives 0 = [1st]:
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│5th│4th│3rd│2nd│ 1■│0th│▒
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│ 2 │ │ 1 │ │ │ │▒
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Step 3 │ 2 │ │ 1 │ │ 0 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Now we consider where to place 5. By (2), it is at 2nd or 0th. If it is at 0th, then we would fail to avoid ⛔「⟨⋯ Perm(1,3,4) ⋯⟩」. Therefore, 5 = [2nd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
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│ 2 │ │ 1 │ │ 0 │ │▒
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Step 4 │ 2 │ │ 1 │ 5 │ 0 │ │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ▧ ▧ − ▢ − − ⟩, Σ▧ ≤ ▢」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│ 0■│▒
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│ 2 │ │ 1 │ 5 │ 0 │ │▒
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Step 5 │ 2 │ 4 │ 1 │ 5 │ 0 │ │▒
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Step 6 │ 2 │ 4 │ 1 │ 5 │ 0 │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9