There is a row with $2024$ cells. Ana and Beto take turns playing, with Ana going first. On each turn, the player selects an empty cell and places a digit in that space. Once all $2024$ cells are filled, the number obtained from reading left to right is considered, ignoring any leading zeros. Beto wins if the resulting number is a multiple of $99$, otherwise Ana wins. Determine which of the two players has a winning strategy and describe it.
Problem
Source: Centroamerican and Caribbean Math Olympiad 2024 P2
Tags: OMCC, combinatorics, winning strategy
Supertinito
22.10.2024 17:29
This problem was proposed by Christian Gonzalez and myself (Santiago Rodriguez). Both of us from Colombia. I hope that you enjoy it.
starchan
22.10.2024 18:45
Pair up $(x, x+2)$ for all $x \equiv 0, 1 \pmod 4$ positions. When Ana plays $i$ in a position, Beto plays $9-i$ in the position its paired with. It may be checked that this wins for Beto.
UglyScientist
30.11.2024 16:02
This pretty looks like ISL 2017 N2 but the roles are changed. My solution is same as #3. Let the entries be $a_1, a_2,...a_{2024}$ looking at residue we see that Beto wants $a_1+a_2+...+a_n$ to be divisible by 9 (1) and $a_2-a_1+a_4-a_3+...+a_{2024}-a_{2023}$ to be divisible by 11 (2). So we can just divide the index to odds and evens. When Ana choose an index and number $d$ for that index than Beto can choose $9-d$ corresponding to an index with the same parity. Since Beto makes the even numbered moves both odd and even indexes will be end after Beto's move. Now this strategy satisfies-(1) is obvious because every pair sum is divisible by 9 and (2) will be true because $11|9 \cdot 1012$.