Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
Sim⟨ ⁵ᵗʰ1 ⁴ᵗʰ0 ³ʳᵈ5 ²ⁿᵈ3 ¹ˢᵗ4 ⁰ᵗʰ2 ⟩ ≥ 2
⟨⋯ 2 ⋯ 3 ⋯ 5 ⋯⟩
⛔Avoid
⟨⋯ ᵃb ⋯⟩, ab=5
min {p5, p4} = 0
{p5, p2, p0} = ? + {0,1,2}
#125034_v2.13
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 3 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ │ 3 │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ │ 0 │ 3 │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 0 │ 3 │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 0 │ 3 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2026-05-19 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
To match ✅「Sim⟨ ⁵ᵗʰ1 ⁴ᵗʰ0 ³ʳᵈ5 ²ⁿᵈ3 ¹ˢᵗ4 ⁰ᵗʰ2 ⟩ ≥ 2」, we need at least two positional digits agreed with ⟨105342⟩.
By ⛔「⟨⋯ ᵃb ⋯⟩, ab=5」, 1 is not an agreed digit. By ⛔「min {p5, p4} = 0」, neither is 0. By ✅「⟨⋯ 2 ⋯ 3 ⋯ 5 ⋯⟩」, 2 is also not an agreed digit. Therefore,
(1) at least two of the following hold:
• [3rd] = 5
• [2nd] = 3
• [1st] = 4
If 4 is not an agreed digit, then (1) gives
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 5 │ 3 │ │ │
└───┴───┴───┴───┴───┴───┘
and we cannot match ✅「⟨⋯ 2 ⋯ 3 ⋯ 5 ⋯⟩」. Therefore, 4 is indeed an agreed digit.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 4 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
In view of (1), we check what happens if 5 is an agreed digit:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 5 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┘
By ✅「⟨⋯ 2 ⋯ 3 ⋯ 5 ⋯⟩」, we have
┌───┬───┬───┬───┬───┬───┐
│ 5▲│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 3 │ 5 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
It follows that {[5th], [2nd], [0th]} = {0,1,2}. In view of ⛔「{p5, p2, p0} = ? + {0,1,2}」, it is a contradiction.
So 5 is not an agreed digit. Using this with (1), the agreed digits are 3 and 4 actually.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 3 │ 4 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Then ✅「⟨⋯ 2 ⋯ 3 ⋯ 5 ⋯⟩」 implies [0th]=5.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 3 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ │ 3 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「min {p5, p4} = 0」, we need 0=[3rd].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 3 │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ │ 0 │ 3 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨⋯ ᵃb ⋯⟩, ab=5」, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5■│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 0 │ 3 │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 0 │ 3 │ 4 │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 0 │ 3 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.13