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.

How can demon technology be prevented from surpassing humans?

Do businesses save their customers' credit card information until the payment is finalized?

Is Having my Players Control Two Parties a Good Idea?

Does the warlock's Gift of the Ever-Living Ones eldritch invocation work with potions or healing spells cast on you by others?

What is the design rationale for having armor and magic penetration mechanics?

Did smallpox emerge in 1580?

QGIS 3.4 calculate date, add month

Is it now possible to undetectably cross the Arctic Ocean on ski/kayak?

What is the German word for: "It only works when I try to show you how it does not work"?

Will the size of Bitcoin core full-node be too big to run on a normal computer?

What is a recommended strategy on exercises in a mathematical textbook at graduate level?

Can it be improved? Help me to fix this code

How does Firefox know my ISP login page?

How can a "proper" function have a vertical slope?

Novel set in the future, children cannot change the class they are born into, one class is made uneducated by associating books with pain

Is there any restriction in entering the South American countries multiple times in one year?

Why exactly is the answer 50 Ohms?

Boot directly into another kernel from running Linux without bootloader

How to protect my Wi-Fi password from being displayed by Android phones when sharing it with QR code?

Skewer removal without quick release

Diamondize Some Text

How should I tell a professor the answer to something he doesn't know?

How can a company compel a W2 employee to sign a non-compete agreement?

What does the British parliament hope to achieve by requesting a third Brexit extension?



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.






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;








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














$begingroup$



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






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.










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.








share|cite|improve this question




share|cite|improve this question



share|cite|improve this question








edited 9 hours ago









N. F. Taussig

50k10 gold badges37 silver badges60 bronze badges




50k10 gold badges37 silver badges60 bronze badges






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.








asked 10 hours ago









BiobbbBiobbb

141 bronze badge




141 bronze badge




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.




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.







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
2






active

oldest

votes


















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












$endgroup$






















    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

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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

                  25.4k4 gold badges16 silver badges42 bronze badges




                  25.4k4 gold badges16 silver badges42 bronze badges
















                      Popular posts from this blog

                      Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

                      Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

                      Ласкавець круглолистий Зміст Опис | Поширення | Галерея | Примітки | Посилання | Навігаційне меню58171138361-22960890446Bupleurum rotundifoliumEuro+Med PlantbasePlants of the World Online — Kew ScienceGermplasm Resources Information Network (GRIN)Ласкавецькн. VI : Літери Ком — Левиправивши або дописавши її