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2^n-1 | Sigma(2^n_i) => k>=n

Source: Pre-VMO 2012 - Round 2 - Problem 5

December 26, 2011
modular arithmeticnumber theory proposednumber theory

Problem Statement

Let n≥2n \geq 2 be a positive integer. Suppose there exist non-negative integers n1,n2,…,nk{n_1},{n_2},\ldots,{n_k} such that 2n−1∣∑i=1k2ni2^n - 1 \mid \sum_{i = 1}^k {{2^{{n_i}}}}. Prove that k≥nk \ge n.
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