Dgdrxrt

Is flight data recorder erased after every flight?

How do you say "canon" as in "official for a story universe"?

Which Sci-Fi work first showed weapon of galactic-scale mass destruction?

Can't find the latex code for the ⍎ (down tack jot) symbol

Inline version of a function returns different value than non-inline version

Where to refill my bottle in India?

Extreme, unacceptable situation and I can't attend work tomorrow morning

Deadlock Graph and Interpretation, solution to avoid

aging parents with no investments

Dual Citizen. Exited the US on Italian passport recently

Carnot-Caratheodory metric

I looked up a future colleague on LinkedIn before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?

JSON.serialize: is it possible to suppress null values of a map?

Should I write numbers in words or as numerals when there are multiple next to each other?

Why do UK politicians seemingly ignore opinion polls on Brexit?

Is three citations per paragraph excessive for undergraduate research paper?

How to create dashed lines/arrows in Illustrator

Understanding the implication of what "well-defined" means for the operation in quotient group

Is "plugging out" electronic devices an American expression?

Does light intensity oscillate really fast since it is a wave?

Inversion Puzzle

Can the Protection from Evil and Good spell be used on the caster?

How to reverse every other sublist of a list?

How to change the limits of integration

Combinatorics Question (card question) [closed]

1

2

$begingroup$

Suppose we randomly arrange $100$ cards in a circle. If $41$ of the cards are red and $59$ of the cards are black, show that you can always find $2$ red cards such that there are exactly $19$ cards between them.

combinatorics

share|cite|improve this question

asked Mar 30 at 11:58

FC BarcelonaFC Barcelona

61

$endgroup$

closed as off-topic by egreg, NCh, Eevee Trainer, Cesareo, Javi Apr 2 at 12:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – egreg, NCh, Eevee Trainer, Cesareo, Javi

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think your question might be incomplete, please edit it to make it clearer. After that, can you please tell us what have you tried to do to solve this question and what is exactly your difficulty?
  $endgroup$
  – Ertxiem
  Mar 30 at 12:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is complete; what is it that you are unsure of? I tried using pigeonhole principle as suggested by the user below but I can't seem to find how to apply it. Thanks in advance.
  $endgroup$
  – FC Barcelona
  Mar 30 at 13:04
  

  
  
  
  

add a comment | 

1

2

$begingroup$

Suppose we randomly arrange $100$ cards in a circle. If $41$ of the cards are red and $59$ of the cards are black, show that you can always find $2$ red cards such that there are exactly $19$ cards between them.

combinatorics

share|cite|improve this question

asked Mar 30 at 11:58

FC BarcelonaFC Barcelona

61

$endgroup$

closed as off-topic by egreg, NCh, Eevee Trainer, Cesareo, Javi Apr 2 at 12:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – egreg, NCh, Eevee Trainer, Cesareo, Javi

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think your question might be incomplete, please edit it to make it clearer. After that, can you please tell us what have you tried to do to solve this question and what is exactly your difficulty?
  $endgroup$
  – Ertxiem
  Mar 30 at 12:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is complete; what is it that you are unsure of? I tried using pigeonhole principle as suggested by the user below but I can't seem to find how to apply it. Thanks in advance.
  $endgroup$
  – FC Barcelona
  Mar 30 at 13:04
  

  
  
  
  

add a comment | 

$begingroup$

Suppose we randomly arrange $100$ cards in a circle. If $41$ of the cards are red and $59$ of the cards are black, show that you can always find $2$ red cards such that there are exactly $19$ cards between them.

combinatorics

share|cite|improve this question

asked Mar 30 at 11:58

FC BarcelonaFC Barcelona

61

$endgroup$

share|cite|improve this question

asked Mar 30 at 11:58

FC BarcelonaFC Barcelona

61

share|cite|improve this question

asked Mar 30 at 11:58

FC BarcelonaFC Barcelona

61

asked Mar 30 at 11:58

FC BarcelonaFC Barcelona

61

asked Mar 30 at 11:58

FC BarcelonaFC Barcelona

61

1 Answer
1

active

oldest

votes

1

$begingroup$

Hint:

If you number the positions of the $100$ cards from $1$ to $100$, then if you consider $2$ cards that have "exactly $19$ cards between them", that means that the difference of their numbers (higher value - lower value) is either $20$ or $80$ (if the gap between card $100$ and $1$ is between them, e.g. card $95$ and card $15$). In either case, the difference is divisible by $20$.

Considering that $100=5times 20$, and $41=2times20+1$, can you make an argument using the pigeon hole principle twice to prove what you need to prove?

share|cite|improve this answer

answered Mar 30 at 12:43

IngixIngix

5,192159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right. I'm sorry for my mistake. I'll delete my other comment so that it does not confuse others.
  $endgroup$
  – Ertxiem
  Mar 31 at 2:09
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint:

If you number the positions of the $100$ cards from $1$ to $100$, then if you consider $2$ cards that have "exactly $19$ cards between them", that means that the difference of their numbers (higher value - lower value) is either $20$ or $80$ (if the gap between card $100$ and $1$ is between them, e.g. card $95$ and card $15$). In either case, the difference is divisible by $20$.

Considering that $100=5times 20$, and $41=2times20+1$, can you make an argument using the pigeon hole principle twice to prove what you need to prove?

share|cite|improve this answer

answered Mar 30 at 12:43

IngixIngix

5,192159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right. I'm sorry for my mistake. I'll delete my other comment so that it does not confuse others.
  $endgroup$
  – Ertxiem
  Mar 31 at 2:09
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint:

If you number the positions of the $100$ cards from $1$ to $100$, then if you consider $2$ cards that have "exactly $19$ cards between them", that means that the difference of their numbers (higher value - lower value) is either $20$ or $80$ (if the gap between card $100$ and $1$ is between them, e.g. card $95$ and card $15$). In either case, the difference is divisible by $20$.

Considering that $100=5times 20$, and $41=2times20+1$, can you make an argument using the pigeon hole principle twice to prove what you need to prove?

share|cite|improve this answer

answered Mar 30 at 12:43

IngixIngix

5,192159

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right. I'm sorry for my mistake. I'll delete my other comment so that it does not confuse others.
  $endgroup$
  – Ertxiem
  Mar 31 at 2:09
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint:

If you number the positions of the $100$ cards from $1$ to $100$, then if you consider $2$ cards that have "exactly $19$ cards between them", that means that the difference of their numbers (higher value - lower value) is either $20$ or $80$ (if the gap between card $100$ and $1$ is between them, e.g. card $95$ and card $15$). In either case, the difference is divisible by $20$.

Considering that $100=5times 20$, and $41=2times20+1$, can you make an argument using the pigeon hole principle twice to prove what you need to prove?

share|cite|improve this answer

answered Mar 30 at 12:43

IngixIngix

5,192159

$endgroup$

share|cite|improve this answer

answered Mar 30 at 12:43

IngixIngix

5,192159

answered Mar 30 at 12:43

IngixIngix

5,192159

answered Mar 30 at 12:43

IngixIngix

5,192159