Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
4th → a, 2nd → b, a+b=3
⟦1,5⟧ ∋ 2,3
⟨ ³ʳᵈa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 51
6th → a, 1st → b, a+b=1+4n
⛔Avoid
1st → 4
----- Information -----
🔲 「4th → a, 2nd → b, a+b=3」
480 permutations match this pattern.
Examples: ⟨0426135⟩, ⟨2135046⟩, ⟨6204351⟩.
🔲 「⟦1,5⟧ ∋ 2,3」
The closed interval given by 1 and 5 contains 2, 3.
840 permutations match this pattern.
Examples: ⟨0413256⟩, ⟨0532461⟩, ⟨1403625⟩.
🔲 「⟨ ³ʳᵈa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 51」
(ab)₁₀ ∶= 10a + b.
1320 permutations match this pattern.
Examples: ⟨3605241⟩, ⟨4016253⟩, ⟨0216354⟩.
🔲 「6th → a, 1st → b, a+b=1+4n」
[6th] + [1st] = 1 | 5 | 9 | ⋯ .
1440 permutations match this pattern.
Examples: ⟨5021643⟩, ⟨5601243⟩, ⟨6054231⟩.
🔳 「1st → 4」
720 permutations match this pattern.
Examples: ⟨1206345⟩, ⟨2306541⟩, ⟨6132045⟩.
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 2 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 2 │ 6 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 3 │ 2 │ 6 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 3 │ 2 │ 6 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 3 │ 2 │ 6 │ 1 │ │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 5 │ 3 │ 2 │ 6 │ 1 │ 0 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2023-12-18 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「4th → a, 2nd → b, a+b=3」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│*4 │3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ X │ │ Y │ │ │
└───┴───┴───┴───┴───┴───┴───┘
(1) where {X,Y} = {0,3} or {1,2}.
(2) On the other hand, by ✅「⟨ ³ʳᵈa ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 51」, we have [3rd] = 5 or 6. We claim that [3rd] = 6.
------------------------------
It is proved by contradiction. Suppose on the contrary [3rd] = 5.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ X │ 5 │ Y │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Note that ✅「⟦1,5⟧ ∋ 2,3」 implies 1,5 are not adjacent. Therefore, it follows from (1) that {X,Y} = {0,3}. Combining this with the preceding pattern, there are two possibilities:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(3) │ │ │ 0 │ 5 │ 3 │ 2 │ 1 │
├───┼───┼───┼───┼───┼───┼───┤
(4) │ 1 │ 2 │ 3 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
We cannot match ✅「6th → a, 1st → b, a+b=1+4n」 if it is case (3). In case (4), we need [1st] = 4 to match the preceding pattern, but then we match ⛔「1st → 4」 too, which is a contradiction.
------------------------------
Our claim in (2) is verified. Accordingly
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ X │ 6 │ Y │ │ │
└───┴───┴───┴───┴───┴───┴───┘
(5) Now, we consider whether {X,Y} = {0,3} or {1,2}. We claim that the latter holds actually.
------------------------------
It is again proved by contradiction. If on the contrary {X,Y} = {0,3}, then the idle numbers become
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Observe that to match ✅「6th → a, 1st → b, a+b=1+4n」, namely to satisfy
A + B = 1|5|9, A := [6th], B := [1st],
we need {A,B} = {1,4} or {4,5}. A fortiori, 4 has to be the value of [6th] or [1st]. Combining this with ⛔「1st → 4」, we find that
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ │ X │ 6 │ Y │ B │ │
└───┴───┴───┴───┴───┴───┴───┘
where B = 1 or 5.
Recall that {X,Y} = {0,3} by assumption. So, regardless of B = 1 or 5, we cannot match ✅「⟦1,5⟧ ∋ 2,3」. This shows a contradiction.
------------------------------
Our claim in (5) is thus verified. It means that we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ X │ 6 │ Y │ │ │
└───┴───┴───┴───┴───┴───┴───┘
where {X,Y} = {1,2}. Combining this with ✅「⟦1,5⟧ ∋ 2,3」, there are two possibilities:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│*4 │3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(6) │ │ │ 1 │ 6 │ 2 │ 3 │ 5 │
├───┼───┼───┼───┼───┼───┼───┤
(7) │ 5 │ 3 │ 2 │ 6 │ 1 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
We cannot match ✅「6th → a, 1st → b, a+b=1+4n」 if case (6) holds. So case (7) holds instead. In view of ⛔「1st → 4」, we finish by ⟨5326104⟩.
Q.E.D.
#125034_v2.2