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x_1a_1 + ...x_{11}a_{ 11} is divisible by 1989.

Source: Polish MO Recond Round 1989 p4

September 9, 2024
number theorydividesdivisible

Problem Statement

The given integers are a1,a2,…,a11 a_1, a_2, \ldots , a_{11} . Prove that there exists a non-zero sequence x1,x2,…,x11 x_1, x_2, \ldots, x_{11} with terms from the set {−1,0,1} \{-1,0,1\} such that the number x1a1+…x11a11 x_1a_1 + \ldots x_{11}a_{ 11} is divisible by 1989.
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