Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ 0 ⋯ 2 ⋯ 3 ⋯⟩
⟦4,5⟧ ∋ 1
⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯⟩
⟨⋯ ? ⋯ 6 ⋯ (?−1)⟩ (?≠6)
⛔Avoid
⟨ ⁶ᵗʰa ⁵ᵗʰb ⁴ᵗʰc ⟩, (abc)₁₀ ≤ 163
1st → a, 0th → b, a+b=8
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 5 │ │ 3 │▒
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Step 3 │ │ │ │ │ 5 │ 6 │ 3 │▒
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Step 4 │ │ │ 2 │ │ 5 │ 6 │ 3 │▒
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Step 5 │ │ │ 2 │ 1 │ 5 │ 6 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ │ 2 │ 1 │ 5 │ 6 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 0 │ 2 │ 1 │ 5 │ 6 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2023-12-17 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
(1) We begin with determining the value of [0th]. We have [0th] = 3, because:
(1.1) By ✅「⟨⋯ ? ⋯ 6 ⋯ (?−1)⟩ (?≠6)」, we have [0th] != 6,5;
(1.2) By ✅「⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯⟩」, we have [0th] != 4,2;
(1.3) By ✅「⟦4,5⟧ ∋ 1」, we have [0th] != 1;
(1.4) By ✅「⟨⋯ 0 ⋯ 2 ⋯ 3 ⋯⟩」, we have [0th] != 0.
Accordingly, our first step is:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Observe that ✅「⟨⋯ 0 ⋯ 2 ⋯ 3 ⋯⟩」, ✅「⟦4,5⟧ ∋ 1」, and ✅「⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯⟩」 then imply the following patterns:
(2)
⟨⋯ 0 ⋯ 2 ⋯ 1 ⋯ 3⟩, and
⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯ 5 ⋯ 3⟩.
As a consequence, 5 is at the right of 0,1,2,4, so 5 can only placed at the 2nd or 1st position. Since ⛔「1st → a, 0th → b, a+b=8」 forbids 5 = [1st], we have 5 = [2nd].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 5 │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Now, the places at the left of 5 are occupied by 0,1,2,4, so [1st] = 6.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 5 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ │ │ 5 │ 6 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Back to (2), we need to match:
⟨⋯ 0 ⋯ 2 ⋯ 1 ⋯⟩, and
⟨⋯ 4 ⋯ 2 ⋯ 1 ⋯⟩,
therefore ([4th], [3rd]) = (2,1):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 5 │ 6 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ │ 2 │ │ 5 │ 6 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ │ 2 │ 1 │ 5 │ 6 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ⁶ᵗʰa ⁵ᵗʰb ⁴ᵗʰc ⟩, (abc)₁₀ ≤ 163」, we have [6th] != 0. As a result, we have
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 2 │ 1 │ 5 │ 6 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ │ 2 │ 1 │ 5 │ 6 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 0 │ 2 │ 1 │ 5 │ 6 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2