Prove that in a row of 17 random integers there exist some integers written in succession (next to each other), whose sum is divisible by 17. [duplicate]In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?Let $S=3,4,5,6,7,8,9,10,11,12$. Suppose 6 integers are chosen from S. Must there be 2 integers whose sum is 15?Prove that if $|S| ge 2^n−1 + 1$, then $S$ contains two elements which are disjoint from each other.Let S be a set of n integers. Show that there is a subset of S, sum of whose elements is a multiple of n by pigeon holeProve that any $6$- subset of integers $1…14$ is always the union of two distinct subsets of equal sumProve that given any five integers, there will be three for which the sum of the squares of those integers is divisible by 3.

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Prove that in a row of 17 random integers there exist some integers written in succession (next to each other), whose sum is divisible by 17. [duplicate]


In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?Let $S=3,4,5,6,7,8,9,10,11,12$. Suppose 6 integers are chosen from S. Must there be 2 integers whose sum is 15?Prove that if $|S| ge 2^n−1 + 1$, then $S$ contains two elements which are disjoint from each other.Let S be a set of n integers. Show that there is a subset of S, sum of whose elements is a multiple of n by pigeon holeProve that any $6$- subset of integers $1…14$ is always the union of two distinct subsets of equal sumProve that given any five integers, there will be three for which the sum of the squares of those integers is divisible by 3.






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  • In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?

    1 answer



I presume I have to use the pigeonhole principle here, but so far no luck.










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marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    @Dzoooks The OP says that the integers are random.
    $endgroup$
    – saulspatz
    10 hours ago










  • $begingroup$
    @Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
    $endgroup$
    – saulspatz
    10 hours ago










  • $begingroup$
    @Dzoooks Read the first sentence of my last comment again. Or look at my answer.
    $endgroup$
    – saulspatz
    10 hours ago

















2














$begingroup$



This question already has an answer here:



  • In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?

    1 answer



I presume I have to use the pigeonhole principle here, but so far no luck.










share|cite|improve this question









New contributor



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






$endgroup$






marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    @Dzoooks The OP says that the integers are random.
    $endgroup$
    – saulspatz
    10 hours ago










  • $begingroup$
    @Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
    $endgroup$
    – saulspatz
    10 hours ago










  • $begingroup$
    @Dzoooks Read the first sentence of my last comment again. Or look at my answer.
    $endgroup$
    – saulspatz
    10 hours ago













2












2








2





$begingroup$



This question already has an answer here:



  • In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?

    1 answer



I presume I have to use the pigeonhole principle here, but so far no luck.










share|cite|improve this question









New contributor



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






$endgroup$





This question already has an answer here:



  • In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?

    1 answer



I presume I have to use the pigeonhole principle here, but so far no luck.





This question already has an answer here:



  • In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?

    1 answer







discrete-mathematics






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share|cite|improve this question









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Biobbb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 9 hours ago









N. F. Taussig

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asked 10 hours ago









BiobbbBiobbb

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marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 1




    $begingroup$
    @Dzoooks The OP says that the integers are random.
    $endgroup$
    – saulspatz
    10 hours ago










  • $begingroup$
    @Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
    $endgroup$
    – saulspatz
    10 hours ago










  • $begingroup$
    @Dzoooks Read the first sentence of my last comment again. Or look at my answer.
    $endgroup$
    – saulspatz
    10 hours ago












  • 1




    $begingroup$
    @Dzoooks The OP says that the integers are random.
    $endgroup$
    – saulspatz
    10 hours ago










  • $begingroup$
    @Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
    $endgroup$
    – saulspatz
    10 hours ago










  • $begingroup$
    @Dzoooks Read the first sentence of my last comment again. Or look at my answer.
    $endgroup$
    – saulspatz
    10 hours ago







1




1




$begingroup$
@Dzoooks The OP says that the integers are random.
$endgroup$
– saulspatz
10 hours ago




$begingroup$
@Dzoooks The OP says that the integers are random.
$endgroup$
– saulspatz
10 hours ago












$begingroup$
@Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
$endgroup$
– saulspatz
10 hours ago




$begingroup$
@Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
$endgroup$
– saulspatz
10 hours ago












$begingroup$
@Dzoooks Read the first sentence of my last comment again. Or look at my answer.
$endgroup$
– saulspatz
10 hours ago




$begingroup$
@Dzoooks Read the first sentence of my last comment again. Or look at my answer.
$endgroup$
– saulspatz
10 hours ago










2 Answers
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$begingroup$

Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$



If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.






share|cite|improve this answer












$endgroup$






















    1
















    $begingroup$

    HINTS



    Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.






    share|cite|improve this answer












    $endgroup$






















      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

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      3
















      $begingroup$

      Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$



      If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.






      share|cite|improve this answer












      $endgroup$



















        3
















        $begingroup$

        Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$



        If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.






        share|cite|improve this answer












        $endgroup$

















          3














          3










          3







          $begingroup$

          Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$



          If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.






          share|cite|improve this answer












          $endgroup$



          Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$



          If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.







          share|cite|improve this answer















          share|cite|improve this answer




          share|cite|improve this answer



          share|cite|improve this answer








          edited 10 hours ago

























          answered 10 hours ago









          Mohammad Riazi-KermaniMohammad Riazi-Kermani

          54.7k4 gold badges27 silver badges74 bronze badges




          54.7k4 gold badges27 silver badges74 bronze badges


























              1
















              $begingroup$

              HINTS



              Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.






              share|cite|improve this answer












              $endgroup$



















                1
















                $begingroup$

                HINTS



                Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.






                share|cite|improve this answer












                $endgroup$

















                  1














                  1










                  1







                  $begingroup$

                  HINTS



                  Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.






                  share|cite|improve this answer












                  $endgroup$



                  HINTS



                  Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.







                  share|cite|improve this answer















                  share|cite|improve this answer




                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 9 hours ago









                  MJD

                  49k31 gold badges219 silver badges408 bronze badges




                  49k31 gold badges219 silver badges408 bronze badges










                  answered 10 hours ago









                  saulspatzsaulspatz

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