Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?












1












$begingroup$


Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?

I'm not really sure how to even approach this question.

Source: Washington's Monthly Math Hour, 2014






discrete-mathematics intuition pigeonhole-principle







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  • $begingroup$
    I cant even find two.
    $endgroup$
    – Rudi_Birnbaum
    1 hour ago






  • 1




    $begingroup$
    @Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
    $endgroup$
    – JMoravitz
    1 hour ago










  • $begingroup$
    Well, at least I can find three… $1+2+3$
    $endgroup$
    – Wolfgang Kais
    1 hour ago










  • $begingroup$
    @JMoravitz humor?
    $endgroup$
    – Rudi_Birnbaum
    1 hour ago
















1












$begingroup$


Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?

I'm not really sure how to even approach this question.

Source: Washington's Monthly Math Hour, 2014






discrete-mathematics intuition pigeonhole-principle







share|cite|improve this question







New contributor




Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I cant even find two.
    $endgroup$
    – Rudi_Birnbaum
    1 hour ago






  • 1




    $begingroup$
    @Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
    $endgroup$
    – JMoravitz
    1 hour ago










  • $begingroup$
    Well, at least I can find three… $1+2+3$
    $endgroup$
    – Wolfgang Kais
    1 hour ago










  • $begingroup$
    @JMoravitz humor?
    $endgroup$
    – Rudi_Birnbaum
    1 hour ago














1












1








1





$begingroup$


Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?

I'm not really sure how to even approach this question.

Source: Washington's Monthly Math Hour, 2014






discrete-mathematics intuition pigeonhole-principle







share|cite|improve this question







New contributor




Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?

I'm not really sure how to even approach this question.

Source: Washington's Monthly Math Hour, 2014






discrete-mathematics intuition pigeonhole-principle




discrete-mathematics intuition pigeonhole-principle






share|cite|improve this question







New contributor




Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago









Arvin DingArvin Ding

83




83




New contributor




Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    I cant even find two.
    $endgroup$
    – Rudi_Birnbaum
    1 hour ago






  • 1




    $begingroup$
    @Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
    $endgroup$
    – JMoravitz
    1 hour ago










  • $begingroup$
    Well, at least I can find three… $1+2+3$
    $endgroup$
    – Wolfgang Kais
    1 hour ago










  • $begingroup$
    @JMoravitz humor?
    $endgroup$
    – Rudi_Birnbaum
    1 hour ago


















  • $begingroup$
    I cant even find two.
    $endgroup$
    – Rudi_Birnbaum
    1 hour ago






  • 1




    $begingroup$
    @Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
    $endgroup$
    – JMoravitz
    1 hour ago










  • $begingroup$
    Well, at least I can find three… $1+2+3$
    $endgroup$
    – Wolfgang Kais
    1 hour ago










  • $begingroup$
    @JMoravitz humor?
    $endgroup$
    – Rudi_Birnbaum
    1 hour ago
















$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
1 hour ago




$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
1 hour ago




1




1




$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
1 hour ago




$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
1 hour ago












$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
1 hour ago




$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
1 hour ago












$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
1 hour ago




$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
1 hour ago










2 Answers
2






active

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4












$begingroup$

Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$




So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$




$~$




Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.







share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    $$begin{align}
    1+2+3&=6\
    1+2+3+6&=12\
    1+2+3+6+12&=24\
    vdots
    end{align}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
      $endgroup$
      – Ross Millikan
      1 hour ago













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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    active

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    4












    $begingroup$

    Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$




    So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$




    $~$




    Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.







    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$




      So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$




      $~$




      Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.







      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$




        So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$




        $~$




        Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.







        share|cite|improve this answer









        $endgroup$



        Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$




        So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$




        $~$




        Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = sumlimits_{k=1}^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_{k=1}^nx_k$ for all $ileq n$ for all $ngeq 3$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        JMoravitzJMoravitz

        48.2k33886




        48.2k33886























            5












            $begingroup$

            $$begin{align}
            1+2+3&=6\
            1+2+3+6&=12\
            1+2+3+6+12&=24\
            vdots
            end{align}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
              $endgroup$
              – Ross Millikan
              1 hour ago


















            5












            $begingroup$

            $$begin{align}
            1+2+3&=6\
            1+2+3+6&=12\
            1+2+3+6+12&=24\
            vdots
            end{align}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
              $endgroup$
              – Ross Millikan
              1 hour ago
















            5












            5








            5





            $begingroup$

            $$begin{align}
            1+2+3&=6\
            1+2+3+6&=12\
            1+2+3+6+12&=24\
            vdots
            end{align}$$






            share|cite|improve this answer









            $endgroup$



            $$begin{align}
            1+2+3&=6\
            1+2+3+6&=12\
            1+2+3+6+12&=24\
            vdots
            end{align}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            saulspatzsaulspatz

            17k31434




            17k31434












            • $begingroup$
              We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
              $endgroup$
              – Ross Millikan
              1 hour ago




















            • $begingroup$
              We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
              $endgroup$
              – Ross Millikan
              1 hour ago


















            $begingroup$
            We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
            $endgroup$
            – Ross Millikan
            1 hour ago






            $begingroup$
            We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
            $endgroup$
            – Ross Millikan
            1 hour ago












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