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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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3

$begingroup$

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.

discrete-mathematics

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edited 7 hours ago

N. F. Taussig

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asked 8 hours ago

BiobbbBiobbb

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marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 20 mins 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
  8 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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment
 | 

3

$begingroup$

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.

discrete-mathematics

share|cite|improve this question

edited 7 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

asked 8 hours ago

BiobbbBiobbb

1611 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.

$endgroup$

marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 20 mins 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
  8 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
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment
 | 

$begingroup$

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.

discrete-mathematics

share|cite|improve this question

edited 7 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

asked 8 hours ago

BiobbbBiobbb

1611 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.

$endgroup$

share|cite|improve this question

edited 7 hours ago

N. F. Taussig

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asked 8 hours ago

BiobbbBiobbb

1611 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.

share|cite|improve this question

edited 7 hours ago

N. F. Taussig

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asked 8 hours ago

BiobbbBiobbb

1611 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.

edited 7 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

edited 7 hours ago

N. F. Taussig

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asked 8 hours ago

BiobbbBiobbb

1611 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.

asked 8 hours ago

BiobbbBiobbb

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2 Answers
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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

edited 8 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

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$endgroup$

add a comment
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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 $s_jequiv s_kpmod{17}$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.

share|cite|improve this answer

edited 7 hours ago

MJD

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answered 8 hours ago

saulspatzsaulspatz

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$endgroup$

add a comment
 | 

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

edited 8 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.6k44 gold badges2727 silver badges7474 bronze badges

$endgroup$

add a comment
 | 

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

edited 8 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.6k44 gold badges2727 silver badges7474 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

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

edited 8 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.6k44 gold badges2727 silver badges7474 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.6k44 gold badges2727 silver badges7474 bronze badges

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.6k44 gold badges2727 silver badges7474 bronze badges

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.6k44 gold badges2727 silver badges7474 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 $s_jequiv s_kpmod{17}$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.

share|cite|improve this answer

edited 7 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 8 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges

$endgroup$

add a comment
 | 

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 $s_jequiv s_kpmod{17}$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.

share|cite|improve this answer

edited 7 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 8 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

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 $s_jequiv s_kpmod{17}$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.

share|cite|improve this answer

edited 7 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 8 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 8 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges

edited 7 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

edited 7 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 8 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges

answered 8 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges