2026-03-31 WR

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

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

⟨   ⁴ᵗʰa ³ʳᵈb ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 135

⟦4,5⟧ ∋ 1,2

⟨Perm(0,2,5) Perm(1,3,4)⟩

2nd|0th → 3

⛔Avoid

⟨⋯ 4 ⋯ a ⋯⟩, a = 2|3|5

#125034_v2.12

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

Proof of 2026-03-31 WR
══════════════════════

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

To match ✅「⟨   ⁴ᵗʰa ³ʳᵈb ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 135」, we need

(1) [4th] = 0 or 1.

As ✅「⟨Perm(0,2,5) Perm(1,3,4)⟩」 requires

(2) {[5th], [4th], [3rd]} = {0, 2, 5},

we have [4th] = 0:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │   │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Then, since ✅「⟦4,5⟧ ∋ 1,2」 implies 2 is not in the left corner, it follows from (2) that [5th] = 5 and [3rd] = 2:

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

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

To match ✅「2nd|0th → 3」, we need 3 = [2nd] or [0th]. If it is [0th] then we would fail to avoid ⛔「⟨⋯ 4 ⋯ a ⋯⟩, a = 2|3|5」. Therefore, 3 = [2nd]:

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

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

Finally, in view of ✅「⟦4,5⟧ ∋ 1,2」, we finish by

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

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

Q.E.D.

#125034_v2.12