Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁵ᵗʰa ⁴ᵗʰb ³ʳᵈc ⟩, (abc)₁₀ ≤ 201
⟨ ³ʳᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 314
3rd → a, 2nd → b, ab=0+5n
4th → a, 3rd → b, a+b=6
⛔Avoid
⟨ ▧ ▧ − ▢ ▢ − ⟩, Σ▧ = Σ▢
⟨ ⁴ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 42
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 4 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 4 │ 2 │ │ │ │▒
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Step 3 │ │ 4 │ 2 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 4 │ 2 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 2 │ 0 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 2 │ 0 │ 3 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-07-08 WR
══════════════════════
By ✅「4th → a, 3rd → b, a+b=6」, we have
(1) {[4th], [3rd]} = {1,5} or {2,4}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
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│ │ ▬ │ ▬ │ │ │ │
└───┴───┴───┴───┴───┴───┘
We show that the latter holds actually.
------------------------------
If on the contrary {[4th], [3rd]} = {1,5}, then 5 = [4th] or [3rd]. But
(2.1) if 5 = [4th], then we would fail to avoid ⛔「⟨ ⁴ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 42」;
(2.2) else if 5 = [3rd], then we would fail to match ✅「⟨ ³ʳᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 314」.
------------------------------
We have verified that {[4th], [3rd]} = {2,4}. Combining this with ✅「⟨ ³ʳᵈa ¹ˢᵗb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 314」, we have 4 = [4th] and 2 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│ 3■│2nd│1st│0th│▒
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Step 1 │ │ 4 │ │ │ │ │▒
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Step 2 │ │ 4 │ 2 │ │ │ │▒
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--- Idle ---
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│ 1 │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Next, we consider how to match ✅「3rd → a, 2nd → b, ab=0+5n」. Observe that we need [2nd] = 0 or 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
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│ │ 4 │ 2 │ │ │ │
└───┴───┴───┴───┴───┴───┘
If it is 5 then we would fail to avoid ⛔「⟨ ⁴ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≥ 42」. Therefore, [2nd] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 4 │ 2 │ │ │ │▒
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Step 3 │ │ 4 │ 2 │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Given that 0 and 2 have been used, we need [5th] = 1 in order to match ✅「⟨ ⁵ᵗʰa ⁴ᵗʰb ³ʳᵈc ⟩, (abc)₁₀ ≤ 201」:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 4 │ 2 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 4 │ 2 │ 0 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ▧ ▧ − ▢ ▢ − ⟩, Σ▧ = Σ▢」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 4 │ 2 │ 0 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 2 │ 0 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 2 │ 0 │ 3 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9