2024-07-02 Q1(m=6)

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

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨⋯ 0 ⋯ 6 ⋯ 5 ⋯⟩

6th → a, 4th → b, a+b=1+4n

⟨⋯ 3 ⋯ ? ⋯ 1 (?−2)⟩ (?≠3)

⟨⋯ 0 ⋯ 2 ⋯ 6 ⋯ 1 ⋯⟩

#125034_v2.5

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

Proof of 2024-07-02 Q1(m=6)
═══════════════════════════

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

Plainly, our first step follows from ✅「⟨⋯ 3 ⋯ ? ⋯ 1 (?−2)⟩ (?≠3)」.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, we consider the value of [0th]. Note that

(1) it is not 0,2,6 by ✅「⟨⋯ 0 ⋯ 2 ⋯ 6 ⋯ 1 ⋯⟩」;

(2) it is not 3,5 by ✅「⟨⋯ 3 ⋯ ? ⋯ 1 (?−2)⟩ (?≠3)」;

Therefore, we have [0th] = 4.

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

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │ 0 │   │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Then, in view of ✅「⟨⋯ 0 ⋯ 2 ⋯ 6 ⋯ 1 ⋯⟩」, ✅「⟨⋯ 0 ⋯ 6 ⋯ 5 ⋯⟩」, and ✅「⟨⋯ 3 ⋯ ? ⋯ 1 (?−2)⟩ (?≠3)」, we need to match the following patterns:

(3) ⟨⋯ 0 ⋯ 2 ⋯ 6 ⋯ 5 ⋯ 14⟩;

(4) ⟨⋯ 3 ⋯ 6 ⋯ 5 ⋯ 14⟩.

In particular, there are four digits (0,2,3,6) at the left of 5, and three digits (0,2,3) at the left of 6. Accordingly, we get

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

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

Then, to match ✅「6th → a, 4th → b, a+b=1+4n」, we need

(5) {[6th], [4th]} = {2,3}.

In view of ✅「⟨⋯ 0 ⋯ 2 ⋯ 6 ⋯ 1 ⋯⟩」, we cannot place 2 at the left corner. Therefore, we have [6th] = 3 and [4th] = 2. We finish by

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

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

Q.E.D.

#125034_v2.5