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Let $a_1,a_2,...,a_{11}$ be integers.

Prove that there are numbers $b_1,b_2,...,b_{11}$,

each $b_i$ equal $-1,0$ or $1$, but not all being $0$, such that the number

$N = a_1b_1 +a_2b_2 +···+a_{11}b_{11}$ is divisible by $2015$.

The question sems so ambiguous, I don't even know where to start. I'd appreciate any hints or a full solution.

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  • $\begingroup$ this is not question 1 on the PAMO 2015. I checked. Question one - Day 1 was an inequality. Question 1 - Day 2 was a question about sum of divisor function $\endgroup$ Jan 29, 2017 at 3:21
  • $\begingroup$ Sorry it's question 3 $\endgroup$
    – user346756
    Jan 29, 2017 at 3:27

1 Answer 1

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Consider all the sums $\sum_{i=1}^{11}a_ix_i$, where $x_i \in \{0,1\}$. There are $2^{11} > 2015$ of these sums, so two of them have the same remainder modulo $2015$. Then we take the difference, which has the required form.

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