Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ a ⋯ 3 ⋯⟩, a = 4|6
⟨? 6 ⋯ (?−3) ⋯ 0 ⋯⟩ (?≠6)
1st → 0|1|2|3|6
⟨ ⁴ᵗʰa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 461
⟦2,6⟧ ∋ 0
⛔Avoid
1st → 1|2|3
⟨⋯ 1 ⋯ ? 5 ⋯ (?−1)⟩ (?≠5,6)
6th → a, 3rd → b, ab=0+5n
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ │ 6 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 2 │ 4 │ 6 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 3 │ 4 │ 6 │ │ │ │ 0 │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 4 │ 4 │ 6 │ │ │ │ 0 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 5 │ 4 │ 6 │ │ │ 5 │ 0 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 6 │ 4 │ 6 │ 3 │ │ 5 │ 0 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 7 │ 4 │ 6 │ 3 │ 1 │ 5 │ 0 │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ Proof of 2024-01-09 Q1(m=6) ═══════════════════════════ Notation: if nth -> a, then we write [nth] = a. Plainly, by ✅「⟨? 6 ⋯ (?−3) ⋯ 0 ⋯⟩ (?≠6)」, we have [5th] = 6: ┌───┬───┬───┬───┬───┬───┬───┐ │6th│ 5■│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ Step 1 │ │ 6 │ │ │ │ │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ We proceed to determine the value of [6th]. By ✅「⟨? 6 ⋯ (?−3) ⋯ 0 ⋯⟩ (?≠6)」 and ✅「⟨⋯ a ⋯ 3 ⋯⟩, a = 4|6」, we have [6th] = 5|4. To avoid ⛔「6th → a, 3rd → b, ab=0+5n」, [6th] cannot be 5. Therefore, [6th] = 4. ┌───┬───┬───┬───┬───┬───┬───┐ │ 6■│5th│4th│3rd│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ │ 6 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 2 │ 4 │ 6 │ │ │ │ │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ 2 │ │ 3 │ 0 │ │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Next, by combining ✅「1st → 0|1|2|3|6」 with ⛔「1st → 1|2|3」, we see that [1st] = 0|6. Since 6 has been used, we have [1st] = 0. ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│ 1■│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 4 │ 6 │ │ │ │ │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 3 │ 4 │ 6 │ │ │ │ 0 │ │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ 2 │ │ 3 │ │ │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ Then, to match ✅「⟦2,6⟧ ∋ 0」, we have to place 2 at 0th. ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│2nd│1st│ 0■│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 4 │ 6 │ │ │ │ 0 │ │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 4 │ 4 │ 6 │ │ │ │ 0 │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ │ │ 3 │ │ │ 5 │ └───┴───┴───┴───┴───┴───┴───┘ We consider where to place 5. By ✅「⟨ ⁴ᵗʰa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 461」, it is not at 4th. By ⛔「6th → a, 3rd → b, ab=0+5n」, it is not at 3rd. Accordingly, 5 = [2nd]. ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│4th│3rd│ 2■│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 4 │ 6 │ │ │ │ 0 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 5 │ 4 │ 6 │ │ │ 5 │ 0 │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ 1 │ │ │ 3 │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ Finally, there is only one way to avoid ⛔「⟨⋯ 1 ⋯ ? 5 ⋯ (?−1)⟩ (?≠5,6)」. We finish by ┌───┬───┬───┬───┬───┬───┬───┐ │6th│5th│ 4■│ 3■│2nd│1st│0th│▒ ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒ │ 4 │ 6 │ │ │ 5 │ 0 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 6 │ 4 │ 6 │ 3 │ │ 5 │ 0 │ 2 │▒ ├───┼───┼───┼───┼───┼───┼───┤▒ Step 7 │ 4 │ 6 │ 3 │ 1 │ 5 │ 0 │ 2 │▒ └───┴───┴───┴───┴───┴───┴───┘▒ ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒ --- Idle --- ┌───┬───┬───┬───┬───┬───┬───┐ │ │ │ │ │ │ │ │ └───┴───┴───┴───┴───┴───┴───┘ Q.E.D. #125034_v2.2
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