Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
max {p5, p2, p1} = 3
⟨ ⁵ᵗʰa ³ʳᵈb ²ⁿᵈc ⟩, a > b > c
min {p5, p3, p1} = 0
⛔Avoid
⟨⋯ 4 ⋯ 0 ⋯⟩
#125034_v2.15
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ │ │ │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ │ │ 0 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 5 │ │ │ 0 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 5 │ 2 │ │ 0 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 5 │ 2 │ 1 │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2026-10-06 WR
══════════════════════
Notation: if Nth -> a, then we write pN = a.
We consider where to place 0. By ✅「min {p5, p3, p1} = 0」, there are three possible positions:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ │ ▬ │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「⟨ ⁵ᵗʰa ³ʳᵈb ²ⁿᵈc ⟩, a > b > c」, 0 cannot be p5 or p3. Therefore, 0 = p1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 0 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, in view of ✅「max {p5, p2, p1} = 3」, there are two possible positions for 3:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ │ │ ▬ │ 0 │ │
└───┴───┴───┴───┴───┴───┘
If 3 = p2, then to match ✅「⟨ ⁵ᵗʰa ³ʳᵈb ²ⁿᵈc ⟩, a > b > c」, we need p5 > 3, but then max {p5, p2, p1} ≥ p5 > 3, so we cannot match ✅「max {p5, p2, p1} = 3」. Therefore, 3 = p5 instead:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ │ │ │ 0 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
On the other hand, to avoid ⛔「⟨⋯ 4 ⋯ 0 ⋯⟩」, plainly we need
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ │ 0 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ │ │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ │ │
└───┴───┴───┴───┴───┴───┘
We consider where to place 5. As it is the maximum digit, to match ✅「⟨ ⁵ᵗʰa ³ʳᵈb ²ⁿᵈc ⟩, a > b > c」 it cannot be p3 or p2. So 5 = p4:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ │ 0 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 5 │ │ │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, using ✅「⟨ ⁵ᵗʰa ³ʳᵈb ²ⁿᵈc ⟩, a > b > c」 once more, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 5 │ │ │ 0 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 5 │ 2 │ │ 0 │ 4 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 5 │ 2 │ 1 │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.15