2026-02-10 Q1(v=0)

Rearrange the digits in ⟨125034⟩ to meet the rules below.

⟨ ⁵ᵗʰ▨ ⁴ᵗʰ▨ ³ʳᵈ▨ ²ⁿᵈ▨ ¹ˢᵗ▨ ⁰ᵗʰ▨ ⟩

✅Match

4 ∾ 5

0 ∾ 3

⛔Avoid

3 ∾ 5

⟨     ³ʳᵈa ²ⁿᵈb     ⟩, (ab)₁₀ ≤ 20

#125034_v2.12

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │ 2 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 2 │   │   │ 2 │ 1 │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 3 │   │   │ 2 │ 1 │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 4 │   │   │ 2 │ 1 │ 0 │ 3 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 5 │   │ 5 │ 2 │ 1 │ 0 │ 3 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 5 │ 2 │ 1 │ 0 │ 3 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2026-02-10 Q1(v=0)
═══════════════════════════

Notation: if nth -> a, then we write [nth] = a.

To avoid ⛔「⟨     ³ʳᵈa ²ⁿᵈb     ⟩, (ab)₁₀ ≤ 20」, we need

(1) [3rd] = 2 | 3 | 4 | 5.

On the other hand, to avoid ⛔「3 ∾ 5」 and match ✅「4 ∾ 5」 at the same time, we need

(2.1) 3 and 5 are not in the same circle; and

(2.2) 3 and 4 are not in the same circle.

Applying them to (1), we get [3rd] = 2|3. If [3rd] = 3, then we cannot match ✅「0 ∾ 3」. Therefore, [3rd] = 2:

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│ 3■│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │ 2 │   │   │   │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │   │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Next, we determine what [2nd] is. In view of (2.1) and (2.2), it is not 5 or 4. To match ✅「0 ∾ 3」, it is not 3. To avoid ⛔「⟨     ³ʳᵈa ²ⁿᵈb     ⟩, (ab)₁₀ ≤ 20」, it is not 0. Therefore, [2nd] = 1:

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│ 2■│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
       │   │   │ 2 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 2 │   │   │ 2 │ 1 │   │   │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Using the same reasoning, we have [1st] = 0:

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│2nd│ 1■│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
       │   │   │ 2 │ 1 │   │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 3 │   │   │ 2 │ 1 │ 0 │   │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │ 5 │   │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘

Then, for [0th], in view of (2.1) and (2.2), it has to be 3:

       ┌───┬───┬───┬───┬───┬───┐
       │5th│4th│3rd│2nd│1st│ 0■│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
       │   │   │ 2 │ 1 │ 0 │   │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 4 │   │   │ 2 │ 1 │ 0 │ 3 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │ 5 │   │   │ 4 │
└───┴───┴───┴───┴───┴───┘

Finally, to match ✅「4 ∾ 5」, we finish by

       ┌───┬───┬───┬───┬───┬───┐
       │ 5■│ 4■│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╡▒
       │   │   │ 2 │ 1 │ 0 │ 3 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 5 │   │ 5 │ 2 │ 1 │ 0 │ 3 │▒
       ├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 5 │ 2 │ 1 │ 0 │ 3 │▒
       └───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│   │   │   │   │   │   │
└───┴───┴───┴───┴───┴───┘

Q.E.D.

#125034_v2.12