olymmath

BxMO 2025 Problem 4

    Let a0,a1,…,a10a_{0},a_{1},\ldots ,a_{10} be integers such that, for each i∈{0,1,…,2047}i\in \{0,1,\ldots ,2047\} , there exists a subset S⊆{0,1,…,10}S\subseteq \{0,1,\ldots ,10\} with

    ∑j∈Saj≡i(mod2048).\sum_{j\in S}a_{j}\equiv i\pmod {2048}.

    Show that for each i∈{0,1,…,10}i\in \{0,1,\ldots ,10\} , there is exactly one j∈{0,1,…,10}j\in \{0,1,\ldots ,10\} such that aja_{j} is divisible by 2i2^{i} but not by 2i+12^{i + 1} .

    Note: ∑j∈Saj\sum_{j\in S}a_{j} is the summation notation, for instance, ∑j∈{2,5}aj=a2+a5\sum_{j\in \{2,5\}}a_{j} = a_{2} + a_{5} , while, for the empty set ∅\varnothing , one defines ∑j∈∅aj=0\sum_{j\in \varnothing}a_{j} = 0 .

    Dataset text NuminaMath olympiads. Quoted with source.

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