Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
1st → a, 0th → b, a+b=3+4n
⟨ ³ʳᵈb ²ⁿᵈa ¹ˢᵗc ⟩, a > b > c
⛔Avoid
⟨ ¹ˢᵗa ⁰ᵗʰb ⟩, a > b
⟨ ⁴ᵗʰa ³ʳᵈb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 254
3rd|2nd|1st|0th → 3
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 3 │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 3 │ │ 5 │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 3 │ 4 │ 5 │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-04-08 WR
══════════════════════
By ✅「1st → a, 0th → b, a+b=3+4n」, we have [1st] + [0th] = 3|7. Therefore, for S := {[1st], [0th]}, one of the following holds:
(A) S = {2,5}
(B) S = {3,4}
(C) S = {0,3}
(D) S = {1,2}
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
(1) We show that case (D) holds actually.
------------------------------
Firstly, note that by ⛔「3rd|2nd|1st|0th → 3」, cases (B) and (C) do not hold. Secondly, if (A) holds, then using ✅「⟨ ³ʳᵈb ²ⁿᵈa ¹ˢᵗc ⟩, a > b > c」 and that 5 is not less than any other digit, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ 2 │ 5 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
The preceding required pattern then implies:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 3 │ 4 │ 2 │ 5 │
└───┴───┴───┴───┴───┴───┘
But this matches ⛔「3rd|2nd|1st|0th → 3」, which is a contradiction.
------------------------------
We have verified (1). Accordingly, {[1st], [0th]} = S = {1,2}. To avoid ⛔「⟨ ¹ˢᵗa ⁰ᵗʰb ⟩, a > b」, we need [1st] < [0th]. Therefore, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, we consider where to place 0. As 0 is the minimum digit, so by✅「⟨ ³ʳᵈb ²ⁿᵈa ¹ˢᵗc ⟩, a > b > c」, it is not at 3rd or 2nd. It is not at 4th as well, in view of ⛔「⟨ ⁴ᵗʰa ³ʳᵈb ⁰ᵗʰc ⟩, (abc)₁₀ ≤ 254」. Therefore, 0 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ │ │ │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
It then follows from ⛔「3rd|2nd|1st|0th → 3」 that 3 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 3 │ │ │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, using ✅「⟨ ³ʳᵈb ²ⁿᵈa ¹ˢᵗc ⟩, a > b > c」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 3 │ │ │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 3 │ │ 5 │ 1 │ 2 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 3 │ 4 │ 5 │ 1 │ 2 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8