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a-2006=\sum _{i=1}^{2006} 2^ia_i

Source: 2006 VMEO III Shortlist SL N7 Vietnamese Mathematics e - Olympiad https://artofproblemsolving.com/community/c2461015_vmeo__viet

August 30, 2022
number theory

Problem Statement

Prove that there are only finitely positive integer aa such that a−2006=∑i=120062iaia-2006=\sum\limits_{i=1}^{2006} 2^ia_i with {ai}\{a_i\} as divisors (not necessary distinct) of nn.
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