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Prove that with every given 10 primes $p_1,p_ 2,\ldots,p_{10}$,there always exist 10 number which are not simultaneously equal to $0$, get one of three values: $-1$, $0$, $1$ satisfied that: $\sum\limits_{ i=1}^{10}a_ip_i$ is a multiple of 1000

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    thanks for your comment. i have just edited:). To Gerry Myerson: that is symbol familiar in my country, so sorry, edited of course2012-12-10

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The pigeonhole principle takes care of this. There are $2^{10}-1=1023$ non-empty sums, so two are congruent modulo $1000$, etc., etc.

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    Good. Now *you* can post a complete solution here. After a while, you can even accept your own solution.2012-12-10
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Following the guide of Gerry Myerson, i think here is solution of this problem. First, we'll built 1,0 linear combination of 10 primes. There are $2^{10}$ combination except for one Zero combination, so we get $2^{10}-1=1023$ non zero combination. Then exists two of them congruents modulo 1000, take their subtraction, we have 1,0,-1 linear combination which is multiple of 1000. This is not so difficult problem, and we can replace "prime" hypothesis by "abitrary integer". and the number 1000 can replaced by $1. anyway, make it clear of my mind. thanks to Gerry:)

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    @Schulz:don't let your intention to hypothesis "prime". this solution can apply for any$10$integer^^2012-12-11