Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨ ⁶ᵗʰ▨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
3rd → 0|3|6
⟦0,5⟧ ∋ 1,4,6
⛔Avoid
⟨⋯ ᵃb ⋯⟩, a+b=2
{p5, p2, p1} = ? + {0,1,3}
⟨ ⁴ᵗʰa ³ʳᵈb ²ⁿᵈc ⟩, (abc)₁₀ ≤ 264
⟨⋯ ᵃb ⋯⟩, |a-b|=0
#125034_v2.13
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 3 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ │ 3 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ │ 3 │ 6 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ │ 3 │ 6 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 3 │ 6 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 0 │ 3 │ 6 │ 1 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2026-05-01 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
To match ✅「3rd → 0|3|6」 and avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=0」 at the same time, we need
(1) [3rd] = 0 or 6.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ ▬ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
If it is 0:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 0 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
then we cannot match ✅「⟦0,5⟧ ∋ 1,4,6」. Therefore, we have [3rd] = 6:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 6 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider what [4th] is.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ ▬ │ 6 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
To avoid ⛔「⟨ ⁴ᵗʰa ³ʳᵈb ²ⁿᵈc ⟩, (abc)₁₀ ≤ 264」, we need [4th] ≥ 2. We also have to avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=0」, therefore
(2) [4th] = 2 | 3 | 5.
(2.1) We claim that [4th] = 3 actually.
------------------------------
(3.1) If instead [4th] = 2:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 2 │ 6 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
then ⛔「⟨ ⁴ᵗʰa ³ʳᵈb ²ⁿᵈc ⟩, (abc)₁₀ ≤ 264」 implies [2nd] = 5:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 2 │ 6 │ 5 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
We will fail to match ✅「⟦0,5⟧ ∋ 1,4,6」, which is a contradiction.
(3.2) Else if [4th] = 5:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 5 │ 6 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
then in view of ✅「⟦0,5⟧ ∋ 1,4,6」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 5 │ 6 │ │ │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
We have matched ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=0」 however, which is a contradiction.
------------------------------
We have verified (2.1) and get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 3 │ 6 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider the pattern ⛔「{p5, p2, p1} = ? + {0,1,3}」. To avoid it, we need
(4) {p5, p2, p1} != {1,2,4}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ 3 │ 6 │ ▬ │ ▬ │ │
└───┴───┴───┴───┴───┴───┴───┘
On the other hand, note that ✅「⟦0,5⟧ ∋ 1,4,6」 implies 1,4 are not in corners. Therefore
(5) Both 1 and 4 belong to {p5, p2, p1}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │1 4│ 3 │ 6 │1 4│1 4│ │
└───┴───┴───┴───┴───┴───┴───┘
Combining (5) with (4), we see that 2 is not in {p5, p2, p1}. In other words,
(6) 2 = [6th] or [0th].
In view of ⛔「⟨⋯ ᵃb ⋯⟩, a+b=2」, we have 2 = [6th]:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 3 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ │ 3 │ 6 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
It then follows from ✅「⟦0,5⟧ ∋ 1,4,6」 that
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │0 5│ 3 │ 6 │1 4│1 4│0 5│
└───┴───┴───┴───┴───┴───┴───┘
Using ⛔「⟨⋯ ᵃb ⋯⟩, a+b=2」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ 3 │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ │ 3 │ 6 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ │ 3 │ 6 │ 1 │ 4 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=0」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ 3 │ 6 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 3 │ 6 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 0 │ 3 │ 6 │ 1 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.13