Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ³ʳᵈa ²ⁿᵈb ⟩, (ab)₁₀ ≤ 14
⟦1,6⟧ ∋ 0,3,4,5
⟨⋯ ? ⋯ 3 ⋯ (?+4)⟩
5th → a, 1st → b, ab=3+5n
⛔Avoid
⟨ ⁶ᵗʰa ³ʳᵈb ⟩, max⟦a,b⟧ = 5
{p6, p5, p4, p3} = ? + {0,1,2,4}
#125034_v2.5
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 5 │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ │ 0 │ 5 │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ │ │ 0 │ 5 │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 2 │ │ 0 │ 5 │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ │ 2 │ 3 │ 0 │ 5 │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 2 │ 3 │ 0 │ 5 │ 4 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-06-04 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
To match ✅「⟨⋯ ? ⋯ 3 ⋯ (?+4)⟩」, we need
(1) [0th] = 4|5|6.
Since ✅「⟦1,6⟧ ∋ 0,3,4,5」 implies that 4 and 5 are not in the right corner, (1) gives [0th] = 6.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now, 5 is the maximum idle digit. So, to avoid ⛔「⟨ ⁶ᵗʰa ³ʳᵈb ⟩, max⟦a,b⟧ = 5」, we need:
(2) 5 = [2nd] | [1st].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ / │ / │ / │ │ │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
Observe that if 5 = [1st], then
(3) [5th] * [1st] will be divisible by 5.
This is a contradiction, however, if our answer matches ✅「5th → a, 1st → b, ab=3+5n」 as well. Therefore, 5 != [1st]. So, back to (2), we get 5 = [2nd]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 5 │ │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Next, note that ✅「⟨ ³ʳᵈa ²ⁿᵈb ⟩, (ab)₁₀ ≤ 14」 determines the value of [3rd]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 5 │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ │ 0 │ 5 │ │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
We proceed to check the value of [1st]. There are four idle digits left:
(4) [1st] = 1|2|3|4.
We claim that [1st] = 4 actually.
------------------------------
In fact, note that:
(4.1) If [1st] = 1, then we cannot match ✅「⟦1,6⟧ ∋ 0,3,4,5」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 0 │ 5 │ 1 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
(4.2) Else if [1st] = 2, then we cannot match ✅「⟨⋯ ? ⋯ 3 ⋯ (?+4)⟩」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 0 │ 5 │ 2 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
(4.3) Else if [1st] = 3, then we would match ⛔「{p6, p5, p4, p3} = ? + {0,1,2,4}」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 0 │ 5 │ 3 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
------------------------------
Combining (4), (4.1), (4.2), and (4.3), we have verified that [1st] = 4. Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 0 │ 5 │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ │ │ 0 │ 5 │ 4 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Using ✅「5th → a, 1st → b, ab=3+5n」, we get [5th] = 2 as well:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 0 │ 5 │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 2 │ │ 0 │ 5 │ 4 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of ✅「⟨⋯ ? ⋯ 3 ⋯ (?+4)⟩」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ │ 0 │ 5 │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ │ 2 │ 3 │ 0 │ 5 │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 2 │ 3 │ 0 │ 5 │ 4 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.5