2026-09-08 WR

Rearrange the digits in ⟨125034⟩ to meet the rules below.

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

4th → a, 3rd → b, ab=10

⟨   ⁴ᵗʰa     ¹ˢᵗb   ⟩, min⟦a,b⟧ = 1

5th → a, 1st → b, |a-b|=1

5th|4th|2nd → 2

⛔Avoid

1st → a, 0th → b, |a-b|=2

{p4, p2, p1} = ? + {0,1,2}

#125034_v2.14

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │ 2 │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 2 │   │ 2 │ 5 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 3 │   │ 2 │ 5 │ 1 │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 4 │   │ 2 │ 5 │ 1 │ 4 │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ 1 │ 4 │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 5 │ 1 │ 4 │ 0 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2026-09-08 WR
══════════════════════

Notation: if nth -> a, then we write [nth] = a.

By ✅「4th → a, 3rd → b, ab=10」, we have

(1) {[4th], [3rd]} = {2,5}.

Combining this with ✅「5th|4th|2nd → 2」, we get

       ┌───┬───┬───┬───┬───┬───┐
       │5th│ 4■│ 3■│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │ 2 │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 2 │   │ 2 │ 5 │   │   │   │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │   │   │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Next, we consider which digits [2nd] and [1st] are. By ✅「⟨   ⁴ᵗʰa     ¹ˢᵗb   ⟩, min⟦a,b⟧ = 1」, we see that

(2) 1 is in {[2nd], [1st]} while 0 is not.

Given that 1 ∈ {[2nd], [1st]}, if 3 is also in {[2nd], [1st]} then we would match ⛔「{p4, p2, p1} = ? + {0,1,2}」, which is a contradiction. Therefore, 

(3) (3) 3 is not in {[2nd], [1st]}.

Combining (2) with (3), we have

(4) {[2nd], [1st]} = {1,4}, whence {[5th], [0th]} = {0,3}.

┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│0,3│ 2 │ 5 │1,4│1,4│0,3│
└───┴───┴───┴───┴───┴───┘

We check whether 1 = [2nd] or [1st]. If 1 = [1st]:

┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│0,3│ 2 │ 5 │ 4 │ 1 │0,3│
└───┴───┴───┴───┴───┴───┘

then we cannot match ✅「5th → a, 1st → b, |a-b|=1」 and avoid ⛔「1st → a, 0th → b, |a-b|=2」 at the same time. Therefore, actually 1 = [2nd], and we finish by

       ┌───┬───┬───┬───┬───┬───┐
       │ 5■│4th│3rd│ 2■│ 1■│ 0■│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
       │   │ 2 │ 5 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 3 │   │ 2 │ 5 │ 1 │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 4 │   │ 2 │ 5 │ 1 │ 4 │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ 1 │ 4 │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 5 │ 1 │ 4 │ 0 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │   │   │   │   │
└───┴───┴───┴───┴───┴───┘

Q.E.D.

#125034_v2.14