Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟦1,3⟧ ∋ 0,2,4,6
⟨? ⋯ 5 (?+3) ⋯ 6 ⋯⟩ (?≠2)
⟨⋯ ? ⋯ 5 ⋯ (?+3)⟩ (?≠2)
⛔Avoid
⟨ ⁵ᵗʰc ²ⁿᵈa ⁰ᵗʰb ⟩, a > b > c
#125034_v2.3
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 0 │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 0 │ │ │ 2 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 0 │ 5 │ │ 2 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 0 │ 5 │ 4 │ 2 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 0 │ 5 │ 4 │ 2 │ 6 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-03-05 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟦1,3⟧ ∋ 0,2,4,6」, we see that 4 and 6 are not in the right corner. Hence, the "?" in ✅「⟨⋯ ? ⋯ 5 ⋯ (?+3)⟩ (?≠2)」 is 0, and we have
(1) ⟨⋯ 0 ⋯ 5 ⋯ 3⟩.
Combining (1) with ✅「⟦1,3⟧ ∋ 0,2,4,6」, we have
(2) ⟨⋯ 1 ⋯ 0 ⋯ 5 ⋯ 3⟩,
whence ⟦1,3⟧ ∋ 0,2,4,5,6. As ⟦1,3⟧ contains five digits, we see that 1,3 are in the corners:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ │ │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, using ✅「⟨? ⋯ 5 (?+3) ⋯ 6 ⋯⟩ (?≠2)」, we get
(3) ⟨1 ⋯ 5 4 ⋯ 6 ⋯⟩.
Combining (2) and (3), we get
(4) ⟨1 ⋯ 0 ⋯ 5 4 ⋯ 6 ⋯ 3⟩.
A fortiori, 0 = [5th] | [4th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│*5 │*4 │3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ │ │ │ │ 3 │
└───┴───┴───┴───┴───┴───┴───┘
Actually, we have 0 != [4th], for otherwise we get from (4) that
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ 0 │ 5 │ 4 │ 6 │ 3 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
and we would match ⛔「⟨ ⁵ᵗʰc ²ⁿᵈa ⁰ᵗʰb ⟩, a > b > c」. Therefore, 0 = [5th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 0 │ │ │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now, to avoid ⛔「⟨ ⁵ᵗʰc ²ⁿᵈa ⁰ᵗʰb ⟩, a > b > c」, we need [2nd] < 3. It follows that [2nd] = 2:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 0 │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 0 │ │ │ 2 │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, using (4), we finish by ⟨1054263⟩.
Q.E.D.
#125034_v2.3