Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
⟨⋯ ? ⋯ 2 ⋯ (?−1)⟩ (?≠2,3)
4th → 4|5
⟨⋯ ? ⋯ 5 ⋯ (?−2)⟩ (?≠5)
5th → a, 2nd → b, |a-b|=2
#125034_v2.13
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 4 │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 4 │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 4 │ │ 3 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 2 │ 3 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 2 │ 3 │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-05-26 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
We consider what [0th] is.
• By ✅「⟨⋯ ? ⋯ 5 ⋯ (?−2)⟩ (?≠5)」, it is ≤3 and !=3.
• By ✅「⟨⋯ ? ⋯ 2 ⋯ (?−1)⟩ (?≠2,3)」, it is not 2 or 1.
Therefore, we have [0th]=0.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, using ✅「⟨⋯ ? ⋯ 2 ⋯ (?−1)⟩ (?≠2,3)」 and ✅「⟨⋯ ? ⋯ 5 ⋯ (?−2)⟩ (?≠5)」, we have the following required pattern:
(1) ⟨⋯ 1 ⋯ 2 ⋯ 5 ⋯ 0⟩.
A fortiori, 5!=[4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ / │ │ │ │ 0 │
└───┴───┴───┴───┴───┴───┘
Combining this with ✅「4th → 4|5」, we get [4th]=4.
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 4 │ │ │ │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
To match (1), 1 is [5th] or [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2.1) │ 1 │ 4 │ │ │ │ 0 │
├───┼───┼───┼───┼───┼───┤
(2.2) │ │ 4 │ 1 │ 2 │ 5 │ 0 │
└───┴───┴───┴───┴───┴───┘
If (2.2) holds, then we cannot match ✅「5th → a, 2nd → b, |a-b|=2」. Therefore, (2.1) holds instead.
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 4 │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 4 │ │ │ │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Using ✅「5th → a, 2nd → b, |a-b|=2」, we get [2nd]=3.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 4 │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 4 │ │ 3 │ │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, using (1), we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 4 │ │ 3 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 2 │ 3 │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 2 │ 3 │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.13