Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨? 1 ⋯ (?−2) ⋯ 3 ⋯⟩ (?≠3)
{p5, p1, p0} = ? + {0,1,2}
4th → a, 3rd → b, a+b=1
1st → a, 0th → b, a+b=1+4n
#125034_v2.10
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ 0 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ 1 │ 0 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 1 │ 0 │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 1 │ 0 │ 5 │ 2 │ │▒
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Step 6 │ 4 │ 1 │ 0 │ 5 │ 2 │ 3 │▒
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Proof of 2025-10-14 WR
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Notation: if nth -> a, then we write [nth] = a.
Plainly, by ✅「⟨? 1 ⋯ (?−2) ⋯ 3 ⋯⟩ (?≠3)」, we have [4th] = 1:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
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Step 1 │ │ 1 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, it follows from ✅「4th → a, 3rd → b, a+b=1」 that [3rd] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
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│ │ 1 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ 0 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider ✅「{p5, p1, p0} = ? + {0,1,2}」. It requires that [5th], [1st], [0th] are consecutive digits. Therefore,
(1) {[5th], [1st], [0th]} = {2,3,4} or {3,4,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
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│ ▬ │ 1 │ 0 │ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
On the other hand, to match ✅「1st → a, 0th → b, a+b=1+4n」, we need
(2) {[1st], [0th]} = {2,3} or {4,5}.
Combining (2) with (1), we see that one of the following holds:
(3.1) {[1st], [0th]} = {2,3}, and [5th] = 4;
(3.2) {[1st], [0th]} = {4,5}, and [5th] = 3.
If [5th] = 3, then we cannot match ✅「⟨? 1 ⋯ (?−2) ⋯ 3 ⋯⟩ (?≠3)」. Therefore, (3.1) holds and we get [5th] = 4:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
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│ │ 1 │ 0 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ 1 │ 0 │ │ │ │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Furthermore, as {[1st], [0th]} = {2,3}, it follows that [2nd] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
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│ 4 │ 1 │ 0 │ │ │ │▒
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Step 4 │ 4 │ 1 │ 0 │ 5 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, in view of ✅「⟨? 1 ⋯ (?−2) ⋯ 3 ⋯⟩ (?≠3)」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
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│ 4 │ 1 │ 0 │ 5 │ │ │▒
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Step 5 │ 4 │ 1 │ 0 │ 5 │ 2 │ │▒
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Step 6 │ 4 │ 1 │ 0 │ 5 │ 2 │ 3 │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.10