2024-07-23 Q1(m=6)

Rearrange the digits in ⟨1263045⟩ to meet the rules below.

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ ? ⋯ 1 ⋯ (?+6)⟩

max ⊢0⊣ ≥ 3

3rd → a, 2nd → b, ab=5+6n

⟨⋯ Perm(0,1,2) ⋯⟩

⛔Avoid

⟨           ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 43

#125034_v2.5

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │   │   │   │   │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │   │   │   │ 1 │   │   │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │   │   │   │ 1 │ 5 │   │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │   │   │   │ 1 │ 5 │ 3 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │   │   │ 1 │ 5 │ 3 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 0 │   │ 1 │ 5 │ 3 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 0 │ 2 │ 1 │ 5 │ 3 │ 6 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2024-07-23 Q1(m=6)
═══════════════════════════

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

Plainly, the first step follows from ✅「⟨⋯ ? ⋯ 1 ⋯ (?+6)⟩」:

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

To proceed, observe that ✅「3rd → a, 2nd → b, ab=5+6n」 implies [3rd] * [2nd] = 5. Therefore, {[3rd], [2nd]} = {1,5}.

Actually we have ([3rd], [2nd]) = (1,5). For, if on the contrary ([3rd], [2nd]) = (5,1):

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │   │   │ 5 │ 1 │   │ 6 │
└───┴───┴───┴───┴───┴───┴───┘

then we cannot match ✅「⟨⋯ Perm(0,1,2) ⋯⟩」, which is a contradiction. So we get

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │   │ 3 │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

Then, note that to match ✅「⟨⋯ Perm(0,1,2) ⋯⟩」, we need

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │ ▬ │ ▬ │ 1 │ 5 │   │ 6 │
└───┴───┴───┴───┴───┴───┴───┘

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

where "▬" are occupied by 0,2. As a consequence, {[6th], [1st]} = {3,4}. To avoid ⛔「⟨           ¹ˢᵗa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 43」, we need [1st] != 4. It follows that

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │   │   │ 0 │   │   │
└───┴───┴───┴───┴───┴───┴───┘

Finally, in view of ✅「max ⊢0⊣ ≥ 3」, we finish by

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│ 5■│ 4■│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 4 │   │   │ 1 │ 5 │ 3 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 0 │   │ 1 │ 5 │ 3 │ 6 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 0 │ 2 │ 1 │ 5 │ 3 │ 6 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

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

Q.E.D.

#125034_v2.5