Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩
✅Match
min ⊢5⊣ = 2
⛔Avoid
⟨⋯ ᵃb ⋯⟩, a+b=2
⟨⋯ 3 ⋯ 2 ⋯ 5 ⋯⟩
⟨ ⁵ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 132
⟨ ⁵ᵗʰa ⁴ᵗʰb ⟩, (ab)₁₀ ≥ 30
⟨⋯ ᵃb ⋯⟩, |a-b|=3
#125034_v2.13
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ 1 │ │ │ │ │ │▒
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Step 2 │ 1 │ │ 5 │ │ │ │▒
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Step 3 │ 1 │ 2 │ 5 │ │ │ │▒
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Step 4 │ 1 │ 2 │ 5 │ 4 │ │ │▒
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Step 5 │ 1 │ 2 │ 5 │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 2 │ 5 │ 4 │ 3 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2026-04-28 WR
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To avoid ⛔「⟨ ⁵ᵗʰa ⁴ᵗʰb ⟩, (ab)₁₀ ≥ 30」, we need [5th] ≤ 2. Combining this with ⛔「⟨ ⁵ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 132」, we have
(1) [5th] = 1 or 2.
If it is 2, then we fail to avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=3」. Therefore
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
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Step 1 │ 1 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
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│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider where to place 5.
To match ✅「min ⊢5⊣ = 2」, 5 cannot be adjacent to 1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
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(2.1) │ 1 │ / │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=3」, we cannot have 5 = [2nd] as well.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2.2) │ 1 │ / │ │ / │ │ │
└───┴───┴───┴───┴───┴───┘
If 5 = [0th], then using ✅「min ⊢5⊣ = 2」, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ │ │ 2 │ 5 │
└───┴───┴───┴───┴───┴───┘
and we will match ⛔「⟨⋯ 3 ⋯ 2 ⋯ 5 ⋯⟩」, which is a contradiction. Therefore, 5 != [0th].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2.3) │ 1 │ / │ │ / │ │ / │
└───┴───┴───┴───┴───┴───┘
If 5 = [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ │ │ 5 │ │
└───┴───┴───┴───┴───┴───┘
then to match ✅「min ⊢5⊣ = 2」 and avoid ⛔「⟨⋯ ᵃb ⋯⟩, a+b=2」 at the same time, we need 2 = [2nd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ │ 2 │ 5 │ │
└───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「⟨⋯ 3 ⋯ 2 ⋯ 5 ⋯⟩」, we need
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
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│ 1 │ │ │ 2 │ 5 │ 3 │
└───┴───┴───┴───┴───┴───┘
This however matches ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=3」. It follows that 5 != [1st].
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2.4) │ 1 │ / │ │ / │ / │ / │
└───┴───┴───┴───┴───┴───┘
Accordingly, we have 5 = [3rd]:
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│5th│4th│ 3■│2nd│1st│0th│▒
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│ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ 5 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
By ✅「min ⊢5⊣ = 2」, 2 is adjacent to 5. If 2 = [2nd], then we cannot avoid ⛔「⟨ ⁵ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 132」. Therefore, 2 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 2 │ 5 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Now, to avoid ⛔「⟨ ⁵ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 132」, there are two possibilities:
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│5th│4th│3rd│2nd│1st│0th│
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(3.1) │ 1 │ 2 │ 5 │ 3 │ 4 │ │
├───┼───┼───┼───┼───┼───┤
(3.2) │ 1 │ 2 │ 5 │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
If (3.1) holds, then we match ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=3」. Therefore, (3.2) holds instead.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 2 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 2 │ 5 │ 4 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, using ⛔「⟨⋯ ᵃb ⋯⟩, |a-b|=3」 again, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 2 │ 5 │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 2 │ 5 │ 4 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 2 │ 5 │ 4 │ 3 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.13