Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
⟦2,5⟧ ∋ 4
{p4, p3, p1} = ? + {0,2,3}
5th|4th|2nd|0th → 3
⛔Avoid
⟨⋯ 2 ⋯ a ⋯⟩, a = 0|4|5
4th → a, 0th → b, ab=2+4n
⟨⋯ Perm(0,4,5) ⋯⟩
#125034_v2.15
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 3 │ │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 3 │ 0 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 3 │ 0 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 3 │ 0 │ 4 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 3 │ 0 │ 4 │ 2 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2026-09-29 WR
══════════════════════
By ✅「{p4, p3, p1} = ? + {0,2,3}」, there are three possibilities for the set S := {p4, p3, p1}:
(1) S = {0,2,3} or {1,3,4} or {2,4,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ ▬ │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
(1.1) We show that it is {0,2,3} actually.
------------------------------
(2.1) If instead S = {2,4,5}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │245│245│ │245│ │
└───┴───┴───┴───┴───┴───┘
then to avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|4|5」, we need 2=[1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │4,5│4,5│ │ 2 │ │
└───┴───┴───┴───┴───┴───┘
Consider where to place 0. Note that we cannot avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|4|5」 and ⛔「⟨⋯ Perm(0,4,5) ⋯⟩」 at the same time. This shows a contradiction.
(2.2) Else, suppose S = {1,3,4}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │134│134│ │134│ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Then ✅「5th|4th|2nd|0th → 3」 implies 3=[4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 3 │1,4│ │1,4│ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Consider where to place 2. To avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|4|5」, 2 has to be [0th]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 3 │1,4│ │1,4│ 2 │
└───┴───┴───┴───┴───┴───┘
We have matched ⛔「4th → a, 0th → b, ab=2+4n」 however, which is a contradiction.
------------------------------
We have verified (1.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │023│023│ │023│ │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|4|5」, we need 2 is to the right of at least three other digits. Hence, 2=[1st], which is our first step:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │0,3│0,3│ │ 2 │ │
└───┴───┴───┴───┴───┴───┘
Using ✅「5th|4th|2nd|0th → 3」, we find the positions of 3 and 0 as well:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 3 │ │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 3 │ 0 │ │ 2 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, in view of ✅「⟦2,5⟧ ∋ 4」, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 3 │ 0 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 3 │ 0 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 3 │ 0 │ 4 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 3 │ 0 │ 4 │ 2 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.15