can we find multiple of a set of finite numbers that are in the “middle” of numbers mod a prime












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$newcommandN{mathbb N} newcommandceil[1]{lceil#1rceil}$Let $a_1,dots, a_kin N$ be an arbitrary finite set of positive integers. Can we find a prime number $p$ such that $p>k$ (preferably $pgg k$) and a natural number $ninN$ such that
$$a_1n,a_2n,dots,a_kn$$
are equivalent to integers in $[p/k,p-p/k]$ mod $p$?



Note: $[a,b]$ here just means a closed interval in the real line.










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  • $begingroup$
    What does [a,b] mean?
    $endgroup$
    – William Elliot
    Jan 26 at 9:45










  • $begingroup$
    a closed interval in the real line
    $endgroup$
    – quantum
    Jan 26 at 10:43


















1












$begingroup$


$newcommandN{mathbb N} newcommandceil[1]{lceil#1rceil}$Let $a_1,dots, a_kin N$ be an arbitrary finite set of positive integers. Can we find a prime number $p$ such that $p>k$ (preferably $pgg k$) and a natural number $ninN$ such that
$$a_1n,a_2n,dots,a_kn$$
are equivalent to integers in $[p/k,p-p/k]$ mod $p$?



Note: $[a,b]$ here just means a closed interval in the real line.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What does [a,b] mean?
    $endgroup$
    – William Elliot
    Jan 26 at 9:45










  • $begingroup$
    a closed interval in the real line
    $endgroup$
    – quantum
    Jan 26 at 10:43
















1












1








1





$begingroup$


$newcommandN{mathbb N} newcommandceil[1]{lceil#1rceil}$Let $a_1,dots, a_kin N$ be an arbitrary finite set of positive integers. Can we find a prime number $p$ such that $p>k$ (preferably $pgg k$) and a natural number $ninN$ such that
$$a_1n,a_2n,dots,a_kn$$
are equivalent to integers in $[p/k,p-p/k]$ mod $p$?



Note: $[a,b]$ here just means a closed interval in the real line.










share|cite|improve this question











$endgroup$




$newcommandN{mathbb N} newcommandceil[1]{lceil#1rceil}$Let $a_1,dots, a_kin N$ be an arbitrary finite set of positive integers. Can we find a prime number $p$ such that $p>k$ (preferably $pgg k$) and a natural number $ninN$ such that
$$a_1n,a_2n,dots,a_kn$$
are equivalent to integers in $[p/k,p-p/k]$ mod $p$?



Note: $[a,b]$ here just means a closed interval in the real line.







combinatorics number-theory elementary-number-theory discrete-mathematics






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edited Jan 26 at 10:49







quantum

















asked Jan 26 at 8:14









quantumquantum

538210




538210












  • $begingroup$
    What does [a,b] mean?
    $endgroup$
    – William Elliot
    Jan 26 at 9:45










  • $begingroup$
    a closed interval in the real line
    $endgroup$
    – quantum
    Jan 26 at 10:43




















  • $begingroup$
    What does [a,b] mean?
    $endgroup$
    – William Elliot
    Jan 26 at 9:45










  • $begingroup$
    a closed interval in the real line
    $endgroup$
    – quantum
    Jan 26 at 10:43


















$begingroup$
What does [a,b] mean?
$endgroup$
– William Elliot
Jan 26 at 9:45




$begingroup$
What does [a,b] mean?
$endgroup$
– William Elliot
Jan 26 at 9:45












$begingroup$
a closed interval in the real line
$endgroup$
– quantum
Jan 26 at 10:43






$begingroup$
a closed interval in the real line
$endgroup$
– quantum
Jan 26 at 10:43












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Not always. Indeed, let $a_i=i$ for each $i$ and $n$ be any natural number distinct from $1$ and $p$.
Since there are $p$ residues modulo $p$ and $k+1$ numbers of the form $ni$ for $i=0,dots, k$, by pigeonhole principle, there exist $0le i<jle k$ and an integer $Delta$ with $|Delta|lefrac p{k+1} $ such that $jnequiv in+Deltapmod p$. Then $0<j-ile k$ and $(j-i)n$ is equivalent mod $p$ to a unique integer in $[0,p-1]$ which is $|Delta|lefrac p{k+1}<frac p{k}$ or $p-|Delta|ge p-frac p{k+1}>p-frac p{k}$.






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    $begingroup$

    Not always. Indeed, let $a_i=i$ for each $i$ and $n$ be any natural number distinct from $1$ and $p$.
    Since there are $p$ residues modulo $p$ and $k+1$ numbers of the form $ni$ for $i=0,dots, k$, by pigeonhole principle, there exist $0le i<jle k$ and an integer $Delta$ with $|Delta|lefrac p{k+1} $ such that $jnequiv in+Deltapmod p$. Then $0<j-ile k$ and $(j-i)n$ is equivalent mod $p$ to a unique integer in $[0,p-1]$ which is $|Delta|lefrac p{k+1}<frac p{k}$ or $p-|Delta|ge p-frac p{k+1}>p-frac p{k}$.






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      0












      $begingroup$

      Not always. Indeed, let $a_i=i$ for each $i$ and $n$ be any natural number distinct from $1$ and $p$.
      Since there are $p$ residues modulo $p$ and $k+1$ numbers of the form $ni$ for $i=0,dots, k$, by pigeonhole principle, there exist $0le i<jle k$ and an integer $Delta$ with $|Delta|lefrac p{k+1} $ such that $jnequiv in+Deltapmod p$. Then $0<j-ile k$ and $(j-i)n$ is equivalent mod $p$ to a unique integer in $[0,p-1]$ which is $|Delta|lefrac p{k+1}<frac p{k}$ or $p-|Delta|ge p-frac p{k+1}>p-frac p{k}$.






      share|cite|improve this answer









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        $begingroup$

        Not always. Indeed, let $a_i=i$ for each $i$ and $n$ be any natural number distinct from $1$ and $p$.
        Since there are $p$ residues modulo $p$ and $k+1$ numbers of the form $ni$ for $i=0,dots, k$, by pigeonhole principle, there exist $0le i<jle k$ and an integer $Delta$ with $|Delta|lefrac p{k+1} $ such that $jnequiv in+Deltapmod p$. Then $0<j-ile k$ and $(j-i)n$ is equivalent mod $p$ to a unique integer in $[0,p-1]$ which is $|Delta|lefrac p{k+1}<frac p{k}$ or $p-|Delta|ge p-frac p{k+1}>p-frac p{k}$.






        share|cite|improve this answer









        $endgroup$



        Not always. Indeed, let $a_i=i$ for each $i$ and $n$ be any natural number distinct from $1$ and $p$.
        Since there are $p$ residues modulo $p$ and $k+1$ numbers of the form $ni$ for $i=0,dots, k$, by pigeonhole principle, there exist $0le i<jle k$ and an integer $Delta$ with $|Delta|lefrac p{k+1} $ such that $jnequiv in+Deltapmod p$. Then $0<j-ile k$ and $(j-i)n$ is equivalent mod $p$ to a unique integer in $[0,p-1]$ which is $|Delta|lefrac p{k+1}<frac p{k}$ or $p-|Delta|ge p-frac p{k+1}>p-frac p{k}$.







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Feb 2 at 7:51









        Alex RavskyAlex Ravsky

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