Mfcttrf

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.

discrete-mathematics

share|cite|improve this question

edited 9 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

asked 10 hours ago

BiobbbBiobbb

1411 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 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
  

  
  
  
  

add a comment
 | 

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.

discrete-mathematics

share|cite|improve this question

edited 9 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

asked 10 hours ago

BiobbbBiobbb

1411 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 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
  

  
  
  
  

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 9 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

asked 10 hours ago

BiobbbBiobbb

1411 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 9 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

asked 10 hours ago

BiobbbBiobbb

1411 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 9 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

asked 10 hours ago

BiobbbBiobbb

1411 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 9 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

edited 9 hours ago

N. F. Taussig

50k1010 gold badges3737 silver badges6060 bronze badges

asked 10 hours ago

BiobbbBiobbb

1411 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 10 hours ago

BiobbbBiobbb

1411 bronze badge

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

edited 10 hours ago

answered 10 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.7k44 gold badges2727 silver badges7474 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 $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.

share|cite|improve this answer

edited 9 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 10 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 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 10 hours ago

answered 10 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.7k44 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 10 hours ago

answered 10 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.7k44 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 10 hours ago

answered 10 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.7k44 gold badges2727 silver badges7474 bronze badges

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.7k44 gold badges2727 silver badges7474 bronze badges

answered 10 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.7k44 gold badges2727 silver badges7474 bronze badges

answered 10 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

54.7k44 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 $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.

share|cite|improve this answer

edited 9 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 10 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 $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.

share|cite|improve this answer

edited 9 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 10 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 $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.

share|cite|improve this answer

edited 9 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 10 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 10 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges

edited 9 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

edited 9 hours ago

MJD

49k3131 gold badges219219 silver badges408408 bronze badges

answered 10 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges

answered 10 hours ago

saulspatzsaulspatz

25.4k44 gold badges1616 silver badges4242 bronze badges