Making a $1,0,-1$ linear commbination of primes a multiple of $1000$ 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

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 $1<a<1023$. anyway, make it clear of my mind. thanks to Gerry:) 
