Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁴ᵗʰa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 52
⟨⋯ ? ⋯ 4 ⋯ (?+1)⟩ (?≠4,3)
⟨⋯ ? ⋯ 5 ⋯ (?−1)⟩ (?≠5,6)
6th → a, 2nd → b, a+b=6
⛔Avoid
6th|3rd|2nd|1st|0th → 1
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ 6 │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 1 │ 6 │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │ 6 │ │ 4 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 6 │ │ 4 │ 5 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 6 │ 0 │ 4 │ 5 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2023-12-16 Q1(m=6)
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Notation: if nth -> a, then we write [nth] = a.
Note that by ✅「⟨ ⁴ᵗʰa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 52」, we have
(1) [4th] = 5|6.
We consider where to place 1. By ⛔「6th|3rd|2nd|1st|0th → 1」, we have 1 = [5th] or [4th]. To match (1), we have 1 != [4th]. Therefore
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 1 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we determine the value of [0th]. As 1 has been used, it follows from ✅「⟨⋯ ? ⋯ 4 ⋯ (?+1)⟩ (?≠4,3)」 and ✅「⟨⋯ ? ⋯ 5 ⋯ (?−1)⟩ (?≠5,6)」 that [0th] = 2 or 3.
(2) We proceed to show that [0th] = 3.
------------------------------
If on the contrary [0th] = 2:
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│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ │ │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
then to match ✅「6th → a, 2nd → b, a+b=6」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│*6 │5th│4th│3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ 1 │ │ │ Y │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
where {X,Y} = {0,6}.
It implies [4th] != 6, so to match (1), we have [4th] = 5:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ 1 │ 5 │ │ Y │ │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
But then we would never match ✅「⟨⋯ ? ⋯ 5 ⋯ (?−1)⟩ (?≠5,6)」, which is a contradiction.
------------------------------
We have verified (2). Accordingly, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now ✅「⟨⋯ ? ⋯ 4 ⋯ (?+1)⟩ (?≠4,3)」 and ✅「⟨⋯ ? ⋯ 5 ⋯ (?−1)⟩ (?≠5,6)」 becomes
⟨⋯ 2 ⋯ 4 ⋯ 3⟩, and
⟨⋯ 4 ⋯ 5 ⋯ 3⟩;
so we have to match the following pattern:
(3) ⟨⋯ 2 ⋯ 4 ⋯ 5 ⋯ 3⟩.
A fortiori, 5 != [4th]. Hence, by (1), we have [4th] = 6.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ 6 │ │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
To match ✅「6th → a, 2nd → b, a+b=6」, we need {[6th], [2nd]} = {2,4}. Combining this with (3), we have
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ 6 │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 1 │ 6 │ │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 1 │ 6 │ │ 4 │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, there is only one way to match (3). We finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 1 │ 6 │ │ 4 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 1 │ 6 │ │ 4 │ 5 │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 1 │ 6 │ 0 │ 4 │ 5 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2