Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁶ᵗʰa ³ʳᵈb ⟩, a > b
4th → a, 3rd → b, ab=2+4n
⟨ ²ⁿᵈa ⁰ᵗʰb ⟩, min⟦a,b⟧ = 2
⟨ ²ⁿᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 510
2nd → a, 1st → b, |a-b|=1
⛔Avoid
⟨ ⁶ᵗʰ↑ ²ⁿᵈ↑ ⁰ᵗʰ↓ ⟩ after 2×⟨←⟩
Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ3 ³ʳᵈ4 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ6 ⟩ ≥ 2
#125034_v2.6
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│6th│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ │ │ 6 │ │ │ │ │▒
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Step 2 │ │ 0 │ 6 │ │ │ │ │▒
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Step 3 │ │ 0 │ 6 │ 1 │ │ │ │▒
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Step 4 │ │ 0 │ 6 │ 1 │ │ │ 2 │▒
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Step 5 │ 3 │ 0 │ 6 │ 1 │ │ │ 2 │▒
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Step 6 │ 3 │ 0 │ 6 │ 1 │ 4 │ │ 2 │▒
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Step 7 │ 3 │ 0 │ 6 │ 1 │ 4 │ 5 │ 2 │▒
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Proof of 2024-09-10 Q1(m=6)
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Notation: if nth -> a, then we write [nth] = a.
Let S := {[2nd], [1st], [0th]}. By ✅「⟨ ²ⁿᵈa ⁰ᵗʰb ⟩, min⟦a,b⟧ = 2」, we have
(1) 2 ∈ S; and
(2) 0,1 ∉ S.
It follows from (1) that 2 ∉ {[4th], [3rd]}. Therefore, ✅「4th → a, 3rd → b, ab=2+4n」 implies
(3) {[4th], [3rd]} = {1,6} | {3,6} | {5,6}.
In particular, 6 ∈ {[4th], [3rd]}. Since 6 is the maximum digit, it is not at 3rd in view of ✅「⟨ ⁶ᵗʰa ³ʳᵈb ⟩, a > b」. So, 6 = [4th]:
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│6th│5th│ 4■│3rd│2nd│1st│0th│▒
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Step 1 │ │ │ 6 │ │ │ │ │▒
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--- Idle ---
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│ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │
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Next, we consider where to place 0. By (2) and (3), it is not at 3rd, 2nd, 1st, or 0th:
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│6th│5th│4th│3rd│2nd│1st│0th│
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│ │ │ 6 │ / │ / │ / │ / │
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On the other hand, as 0 is the minimum digit, it is not at 6th in view of ✅「⟨ ⁶ᵗʰa ³ʳᵈb ⟩, a > b」. Therefore, 0 = [5th]:
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│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 6 │ │ │ │ │▒
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Step 2 │ │ 0 │ 6 │ │ │ │ │▒
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--- Idle ---
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│ 1 │ 2 │ │ 3 │ │ 4 │ 5 │
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We can determine the position of 1 in a similar way. By (2), it is at 6th or 3rd:
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│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 6 │ │ / │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
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│ 1 │ 2 │ │ 3 │ │ 4 │ 5 │
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If 1 = [6th], then we cannot match ✅「⟨ ⁶ᵗʰa ³ʳᵈb ⟩, a > b」. Therefore, 1 = [3rd]:
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│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ 6 │ │ │ │ │▒
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Step 3 │ │ 0 │ 6 │ 1 │ │ │ │▒
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--- Idle ---
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│ │ 2 │ │ 3 │ │ 4 │ 5 │
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Next, we consider ⛔「⟨ ⁶ᵗʰ↑ ²ⁿᵈ↑ ⁰ᵗʰ↓ ⟩ after 2×⟨←⟩」. Given that 0 = [5th] and 6 = [4th], we definitely have a "↓" at 0th and a "↑" at 6th after 2×⟨←⟩. So, to avoid the preceding pattern, we need [2nd] > [0th]. It follows that 2 != [2nd].
In view of ⛔「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ3 ³ʳᵈ4 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ6 ⟩ ≥ 2」, we have 2 != [1st] as well. Therefore, by (1), we get 2 = [0th]:
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│6th│5th│4th│3rd│2nd│1st│ 0■│▒
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│ │ 0 │ 6 │ 1 │ │ │ │▒
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Step 4 │ │ 0 │ 6 │ 1 │ │ │ 2 │▒
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--- Idle ---
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│ │ │ │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Note that ⛔「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ3 ³ʳᵈ4 ²ⁿᵈ1 ¹ˢᵗ2 ⁰ᵗʰ6 ⟩ ≥ 2」 also implies 5 != [6th]. So, 5 ∈ {[2nd], [1st]}. Combining this with ✅「2nd → a, 1st → b, |a-b|=1」, we have
(4) {[2nd], [1st]} = {4,5}.
As a consequence, 3 = [6th]:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
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│ │ 0 │ 6 │ 1 │ │ │ 2 │▒
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Step 5 │ 3 │ 0 │ 6 │ 1 │ │ │ 2 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ 4 │ 5 │
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Finally, in view of ✅「⟨ ²ⁿᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 510」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
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│ 3 │ 0 │ 6 │ 1 │ │ │ 2 │▒
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Step 6 │ 3 │ 0 │ 6 │ 1 │ 4 │ │ 2 │▒
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Step 7 │ 3 │ 0 │ 6 │ 1 │ 4 │ 5 │ 2 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.6