# Wandering over the divisors of a power of ten

Partial progress on an open problem. Vamshi Jandhyala.

> Players multiply by 2 or 5 or divide by 10, never repeating a divisor. The tournament's natural answer fails from 10⁶ on, and the second player beats whole families of openings on every board.

Canonical: https://vamshij.com/research/wandering-over-divisors
Code and Lean proofs: https://github.com/jvvk/mathematics/tree/main/wandering-over-divisors
Paper (PDF): https://github.com/jvvk/mathematics/blob/main/wandering-over-divisors/paper/note.pdf

A referee names a divisor of `10^n`; two players then take turns multiplying the last number named by 2 or by 5, or dividing it by 10, always naming a new divisor of `10^n`; the player who cannot move loses. Writing `2^x 5^y` as the square `(x, y)`, this is a piece on the board `{0, …, n}²` with steps `(1,0)`, `(0,1)` and `(−1,−1)`. The game is problem 20 of the 23rd All-Ukrainian Tournament of Young Mathematicians (2020), posted by Witold on Mathematics Stack Exchange (question 3689425) and by Vepir on MathOverflow (question 363120), where it has had no answer since 2020.

The note proves:

- **Theorem 1.** For `n = 2, 4, 6, 8` the second player wins from 4, 9, 20 and 31 squares. So from a random divisor of `10^6` the first player wins with probability **29/49**, and the natural answer `1 − (k+1)²/(2k+1)²` to the tournament's part (b) is false from `n = 6` on.
- **Lemma 2.** Sweeping an edge: once the piece reaches the right-hand edge with the column below it unused, the second player wins.
- **Theorem 3.** From a diagonal square `(k, k)`, `k < n`, the openings `(1,0)` and `(0,1)` lose; so `(−1,−1)` is the only possible winning opening.
- **Theorem 5.** From a bottom-row square `(a, 0)`, `1 ≤ a ≤ n`, `n ≥ 2`, the opening `(0,1)` loses; so `(1,0)` is the only possible winning opening.
- **Theorem 7.** The corners `(n, 0)`, `(0, n)` and the square `(2, 0)` lose for the first player when `n ≥ 2`, and `(2, 2)` when `n ≥ 3`.

It also records Vepir's conjectured pattern for odd `n` (Conjecture 8, checked on every square up to `10 × 10`, with `3m² − 4m + 5` losing squares on the `2m × 2m` board), and approaches that fail: copying, square weightings (no weighting decides the winner on `5 × 5` or `7 × 7`), and induction over rectangles.

Preprint, 10 October 2026, not peer reviewed. Lemma 2 and Theorems 3, 5 and 7 are formally verified in Lean 4 for every board size (`lean/`, standard axioms only). The exposition has not yet been independently reviewed. The author used an AI tool (Claude, Anthropic) in this work, as described in the note's acknowledgements, and is responsible for its content.