Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
2nd → a, 0th → b, a+b=2+5n
4th → a, 2nd → b, |a-b|=4
4th → a, 3rd → b, a+b=4+5n
⛔Avoid
5th → a, 4th → b, ab=0+5n
⟨⋯ Perm(0,4,5) ⋯⟩
#125034_v2.10
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ │ 5 │ │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 3 │ 5 │ │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 3 │ 5 │ 0 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 1 │ 3 │ 5 │ 0 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-09-09 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「4th → a, 2nd → b, |a-b|=4」, we have
(1) {[4th], [2nd]} = {0,4} or {1,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ ▬ │ │ │
└───┴───┴───┴───┴───┴───┘
On the other hand, to avoid ⛔「5th → a, 4th → b, ab=0+5n」, neither 0 nor 5 is in {[5th], [4th]}. Combining this with (1), we get
(2) ([4th], [2nd]) = (4,0) or (1,5).
(2.1) We show that ([4th], [2nd]) = (1,5) actually .
------------------------------
Suppose on the contrary ([4th], [2nd]) = (4,0):
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「4th → a, 3rd → b, a+b=4+5n」, we need [3rd] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ 5 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
We have matched ⛔「⟨⋯ Perm(0,4,5) ⋯⟩」, which is a contradiction.
------------------------------
We have verified (2.1). Accordingly we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ 5 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, note that there is only one way to match ✅「2nd → a, 0th → b, a+b=2+5n」:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ │ 5 │ │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
and only one way to match ✅「4th → a, 3rd → b, a+b=4+5n」:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ 5 │ │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 3 │ 5 │ │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「5th → a, 4th → b, ab=0+5n」, we finish by
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ 3 │ 5 │ │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 3 │ 5 │ 0 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 1 │ 3 │ 5 │ 0 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.10