Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
6th → a, 0th → b, ab=0
3rd → a, 2nd → b, ab=0+6n
5th → a, 2nd → b, a+b=0+5n
⛔Avoid
⟨ ⁵ᵗʰb ⁴ᵗʰa ⟩, a > b
Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ5 ⁴ᵗʰ2 ³ʳᵈ6 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ4 ⟩ ≥ 1
⟨⋯ 4 ⋯ 1 ⋯ 3 ⋯ 2 ⋯⟩
⟨⋯ 1 ⋯ ? ⋯ 5 (?−1)⟩ (?≠5,6)
⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1|3|4|5
#125034_v2.2
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│6th│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ 0 │ │ │ │ │ │ │▒
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Step 2 │ 0 │ │ │ │ │ │ 2 │▒
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Step 3 │ 0 │ 4 │ │ │ │ │ 2 │▒
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Step 4 │ 0 │ 4 │ │ │ 6 │ │ 2 │▒
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Step 5 │ 0 │ 4 │ │ 1 │ 6 │ │ 2 │▒
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Step 6 │ 0 │ 4 │ 3 │ 1 │ 6 │ │ 2 │▒
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Step 7 │ 0 │ 4 │ 3 │ 1 │ 6 │ 5 │ 2 │▒
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Proof of 2023-12-19 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
To match ✅「6th → a, 0th → b, ab=0」, we need 0 = [6th] or [0th]. We need to avoid ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1|3|4|5」 too, so 0 = [6th].
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ 0 │ │ │ │ │ │ │▒
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--- Idle ---
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│ 1 │ 2 │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
(1) Note that ⛔「⟨⋯ 2 ⋯ a ⋯⟩, a = 0|1|3|4|5」 implies 2 is at the right of 0,1,3,4,5. Therefore, 2 = [1st] or [0th].
Also, note that ✅「5th → a, 2nd → b, a+b=0+5n」 implies
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│6th│*5 │4th│3rd│*2 │1st│0th│
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│ 0 │ A │ │ │ B │ │ │
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(2) where A + B = 5 or 10.
In view of (1), there are two possibilities:
(2.1) {A,B} = {1,4};
(2.2) {A,B} = {4,6}.
We show that case (2.2) holds actually.
------------------------------
If on the contrary case (2.1) holds, then to avoid ⛔「⟨ ⁵ᵗʰb ⁴ᵗʰa ⟩, a > b」, we need (A,B) = (4,1):
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│6th│ 5▲│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 4 │ │ │ 1 │ │ │
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--- Idle ---
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│ │ 2 │ 6 │ 3 │ │ │ 5 │
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Then, to match ✅「3rd → a, 2nd → b, ab=0+6n」, we need [3rd] = 6:
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│6th│5th│4th│ 3▲│2nd│1st│0th│
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│ 0 │ 4 │ │ 6 │ 1 │ │ │
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However, it follows that we would match ⛔「Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ5 ⁴ᵗʰ2 ³ʳᵈ6 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ4 ⟩ ≥ 1」, which is a contradiction.
------------------------------
We have verified that case (2.2) holds. In particular, 6 = [5th] or [2nd]. Combining this with (1), we see that 2 is at the right of all other numbers. Consequently, 2 = [0th].
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│6th│5th│4th│3rd│2nd│1st│ 0■│▒
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│ 0 │ │ │ │ │ │ │▒
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Step 2 │ 0 │ │ │ │ │ │ 2 │▒
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--- Idle ---
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│ 1 │ │ 6 │ 3 │ │ 4 │ 5 │
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Furthermore, as (2.2) holds, we have
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│6th│*5 │4th│3rd│*2 │1st│0th│
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│ 0 │ A │ │ │ B │ │ 2 │
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where (A,B) = (4,6) or (6,4).
(3) We claim that (A,B) = (4,6) actually.
------------------------------
For, if on the contrary (A,B) = (6,4):
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│6th│ 5▲│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 6 │ │ │ 4 │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
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│ 1 │ │ │ 3 │ │ │ 5 │
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then there is only one way to match ✅「3rd → a, 2nd → b, ab=0+6n」:
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│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 6 │ │ 3 │ 4 │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
No matter how we place 1,5, we would match ⛔「Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ5 ⁴ᵗʰ2 ³ʳᵈ6 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ4 ⟩ ≥ 1」 or ⛔「⟨⋯ 1 ⋯ ? ⋯ 5 (?−1)⟩ (?≠5,6)」. This shows a contradiction.
------------------------------
Our claim in (3) is thus verified. Accordingly
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│6th│ 5■│4th│3rd│ 2■│1st│0th│▒
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│ 0 │ │ │ │ │ │ 2 │▒
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Step 3 │ 0 │ 4 │ │ │ │ │ 2 │▒
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Step 4 │ 0 │ 4 │ │ │ 6 │ │ 2 │▒
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--- Idle ---
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│ 1 │ │ │ 3 │ │ │ 5 │
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We consider where to place 1. To avoid ⛔「Sim⟨ ⁶ᵗʰ3 ⁵ᵗʰ5 ⁴ᵗʰ2 ³ʳᵈ6 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ4 ⟩ ≥ 1」, it is not at 1st, and to avoid ⛔「⟨⋯ 4 ⋯ 1 ⋯ 3 ⋯ 2 ⋯⟩」, it is not at 4th. Therefore, 1 = [3rd].
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│6th│5th│4th│ 3■│2nd│1st│0th│▒
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│ 0 │ 4 │ │ │ 6 │ │ 2 │▒
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Step 5 │ 0 │ 4 │ │ 1 │ 6 │ │ 2 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ │ 5 │
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Finally, to avoid ⛔「⟨⋯ 4 ⋯ 1 ⋯ 3 ⋯ 2 ⋯⟩」, we have 3 = [4th]. We finish by
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│6th│5th│ 4■│3rd│2nd│ 1■│0th│▒
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│ 0 │ 4 │ │ 1 │ 6 │ │ 2 │▒
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Step 6 │ 0 │ 4 │ 3 │ 1 │ 6 │ │ 2 │▒
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Step 7 │ 0 │ 4 │ 3 │ 1 │ 6 │ 5 │ 2 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2