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Maybe pigeonhole problem?

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5

$begingroup$

A bird cage could only fit a maximum of $10$ birds. If a house has $21$ bird cage and $100$ birds. A bird cage is considered overpopulated if it fits $4$ or more birds inside it. How many cages (minimum) are overpopulated so that all the birds take place in a cage?

I tried fitting $10$ birds into the first $10$ cages, and it will result of $10$ overpopulated cages, then I reduce the last full cage, and added it to the empty cages down there. It resulted like this

$$
10
10
10
10
10
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
$$
My result is there are $5$ overpopulated bird cages minimum. Am I right? Or is there any other way to do it? Thanks for helping, appreciate your help!

combinatorics discrete-mathematics pigeonhole-principle discrete-optimization

share|cite|improve this question

edited 7 mins ago

greedoid

44k1155109

asked 1 hour ago

GodlixeGodlixe

956

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You put only $98$ birds...
  $endgroup$
  – user289143
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The way you think here is correct altough you missed $2$ birds, adding them into the next cage gives the correct answer
  $endgroup$
  – Vinyl_coat_jawa
  21 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

A bird cage could only fit a maximum of $10$ birds. If a house has $21$ bird cage and $100$ birds. A bird cage is considered overpopulated if it fits $4$ or more birds inside it. How many cages (minimum) are overpopulated so that all the birds take place in a cage?

I tried fitting $10$ birds into the first $10$ cages, and it will result of $10$ overpopulated cages, then I reduce the last full cage, and added it to the empty cages down there. It resulted like this

$$
10
10
10
10
10
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
$$
My result is there are $5$ overpopulated bird cages minimum. Am I right? Or is there any other way to do it? Thanks for helping, appreciate your help!

combinatorics discrete-mathematics pigeonhole-principle discrete-optimization

share|cite|improve this question

edited 7 mins ago

greedoid

44k1155109

asked 1 hour ago

GodlixeGodlixe

956

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You put only $98$ birds...
  $endgroup$
  – user289143
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The way you think here is correct altough you missed $2$ birds, adding them into the next cage gives the correct answer
  $endgroup$
  – Vinyl_coat_jawa
  21 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

A bird cage could only fit a maximum of $10$ birds. If a house has $21$ bird cage and $100$ birds. A bird cage is considered overpopulated if it fits $4$ or more birds inside it. How many cages (minimum) are overpopulated so that all the birds take place in a cage?

I tried fitting $10$ birds into the first $10$ cages, and it will result of $10$ overpopulated cages, then I reduce the last full cage, and added it to the empty cages down there. It resulted like this

$$
10
10
10
10
10
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
$$
My result is there are $5$ overpopulated bird cages minimum. Am I right? Or is there any other way to do it? Thanks for helping, appreciate your help!

combinatorics discrete-mathematics pigeonhole-principle discrete-optimization

share|cite|improve this question

edited 7 mins ago

greedoid

44k1155109

asked 1 hour ago

GodlixeGodlixe

956

$endgroup$

share|cite|improve this question

edited 7 mins ago

greedoid

44k1155109

asked 1 hour ago

GodlixeGodlixe

956

share|cite|improve this question

edited 7 mins ago

greedoid

44k1155109

asked 1 hour ago

GodlixeGodlixe

956

edited 7 mins ago

greedoid

44k1155109

edited 7 mins ago

greedoid

44k1155109

asked 1 hour ago

GodlixeGodlixe

956

asked 1 hour ago

GodlixeGodlixe

956

5 Answers
5

active

oldest

votes

3

$begingroup$

Say we have $n$ cages each with $3$ birds, then the other $21-n$ cages we must fill with $100-3n$ birds. So $$100-3nleq (21-n)cdot 10implies nleq 15$$

$n= 15$ is achiavable: $15cdot 3+5cdot 10+1cdot 5 =100$.

So you must have at least $6$ overpoulated cages.

share|cite|improve this answer

edited 46 mins ago

answered 1 hour ago

greedoidgreedoid

44k1155109

$endgroup$

add a comment | 

1

$begingroup$

If $16$ cages have $3$ or less birds , there are at least $52$ birds remaining which is too much for $5$ additional cages.

If we have $15$ cages with $3$ birds , we have $6$ additional cages, hence enough place for the remaining $55$ birds.

Hence, at least $6$ cages must be overpopulated.

share|cite|improve this answer

answered 1 hour ago

PeterPeter

48k1139133

$endgroup$

add a comment | 

1

$begingroup$

Start putting $3$ birds in each cage... so in total $3 cdot 21= 63$. Then you have $100-63=37$ birds to put and every cage now can contain only $7$ birds, so you need at least $6$ cages (since $5 cdot 7 < 37 < 6 cdot 7$)

share|cite|improve this answer

answered 1 hour ago

user289143user289143

916313

$endgroup$

add a comment | 

0

$begingroup$

I would think the other way around. First put $3$ birds, that is the maximum amount without being overpopulated into the cages. This results in $3cdot 21=63$ birds placed. At this point no cages are overpopulated but we still have $100-63=37$ birds to place. From the text it seems that it doesn´t matter wether we overpopulate by one or five birds per cage (which is definitely not like in real life). So I would start filling the cages up to $10$ with the remaining birds.

Can you figure out the rest?

You have $37$ more birds to place and each cage can hold $7$ more birds, since $37=5cdot 7+2$ you can fill up $5$ cages to full and put the last $2$ birds in a $6$th cage ending up with $6$ overpopulated cages

share|cite|improve this answer

edited 36 mins ago

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

2,8421030

$endgroup$

add a comment | 

0

$begingroup$

Fill all $21$ cages with $3$ birds.

That leaves $100 - 3cdot 21 = 37$ birds loose.

Cram $7$ birds into each of $37/7 = 5 + 2/7$ cages.

Thus $5cdot 10 + 5 + 1cdot 3 = 100$ gives $6$ overpopulated cages.

How you filled the cages does not total to $100$.

share|cite|improve this answer

edited 35 mins ago

Vinyl_coat_jawa

2,8421030

answered 55 mins ago

William ElliotWilliam Elliot

8,2872720

$endgroup$

add a comment | 

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3

$begingroup$

Say we have $n$ cages each with $3$ birds, then the other $21-n$ cages we must fill with $100-3n$ birds. So $$100-3nleq (21-n)cdot 10implies nleq 15$$

$n= 15$ is achiavable: $15cdot 3+5cdot 10+1cdot 5 =100$.

So you must have at least $6$ overpoulated cages.

share|cite|improve this answer

edited 46 mins ago

answered 1 hour ago

greedoidgreedoid

44k1155109

$endgroup$

add a comment | 

3

$begingroup$

Say we have $n$ cages each with $3$ birds, then the other $21-n$ cages we must fill with $100-3n$ birds. So $$100-3nleq (21-n)cdot 10implies nleq 15$$

$n= 15$ is achiavable: $15cdot 3+5cdot 10+1cdot 5 =100$.

So you must have at least $6$ overpoulated cages.

share|cite|improve this answer

edited 46 mins ago

answered 1 hour ago

greedoidgreedoid

44k1155109

$endgroup$

add a comment | 

$begingroup$

Say we have $n$ cages each with $3$ birds, then the other $21-n$ cages we must fill with $100-3n$ birds. So $$100-3nleq (21-n)cdot 10implies nleq 15$$

$n= 15$ is achiavable: $15cdot 3+5cdot 10+1cdot 5 =100$.

So you must have at least $6$ overpoulated cages.

share|cite|improve this answer

edited 46 mins ago

answered 1 hour ago

greedoidgreedoid

44k1155109

$endgroup$

share|cite|improve this answer

edited 46 mins ago

answered 1 hour ago

greedoidgreedoid

44k1155109

answered 1 hour ago

greedoidgreedoid

44k1155109

answered 1 hour ago

greedoidgreedoid

44k1155109

1

$begingroup$

If $16$ cages have $3$ or less birds , there are at least $52$ birds remaining which is too much for $5$ additional cages.

If we have $15$ cages with $3$ birds , we have $6$ additional cages, hence enough place for the remaining $55$ birds.

Hence, at least $6$ cages must be overpopulated.

share|cite|improve this answer

answered 1 hour ago

PeterPeter

48k1139133

$endgroup$

add a comment | 

1

$begingroup$

If $16$ cages have $3$ or less birds , there are at least $52$ birds remaining which is too much for $5$ additional cages.

If we have $15$ cages with $3$ birds , we have $6$ additional cages, hence enough place for the remaining $55$ birds.

Hence, at least $6$ cages must be overpopulated.

share|cite|improve this answer

answered 1 hour ago

PeterPeter

48k1139133

$endgroup$

add a comment | 

$begingroup$

If $16$ cages have $3$ or less birds , there are at least $52$ birds remaining which is too much for $5$ additional cages.

If we have $15$ cages with $3$ birds , we have $6$ additional cages, hence enough place for the remaining $55$ birds.

Hence, at least $6$ cages must be overpopulated.

share|cite|improve this answer

answered 1 hour ago

PeterPeter

48k1139133

$endgroup$

share|cite|improve this answer

answered 1 hour ago

PeterPeter

48k1139133

answered 1 hour ago

PeterPeter

48k1139133

answered 1 hour ago

PeterPeter

48k1139133

1

$begingroup$

Start putting $3$ birds in each cage... so in total $3 cdot 21= 63$. Then you have $100-63=37$ birds to put and every cage now can contain only $7$ birds, so you need at least $6$ cages (since $5 cdot 7 < 37 < 6 cdot 7$)

share|cite|improve this answer

answered 1 hour ago

user289143user289143

916313

$endgroup$

add a comment | 

1

$begingroup$

Start putting $3$ birds in each cage... so in total $3 cdot 21= 63$. Then you have $100-63=37$ birds to put and every cage now can contain only $7$ birds, so you need at least $6$ cages (since $5 cdot 7 < 37 < 6 cdot 7$)

share|cite|improve this answer

answered 1 hour ago

user289143user289143

916313

$endgroup$

add a comment | 

$begingroup$

Start putting $3$ birds in each cage... so in total $3 cdot 21= 63$. Then you have $100-63=37$ birds to put and every cage now can contain only $7$ birds, so you need at least $6$ cages (since $5 cdot 7 < 37 < 6 cdot 7$)

share|cite|improve this answer

answered 1 hour ago

user289143user289143

916313

$endgroup$

share|cite|improve this answer

answered 1 hour ago

user289143user289143

916313

answered 1 hour ago

user289143user289143

916313

answered 1 hour ago

user289143user289143

916313

0

$begingroup$

I would think the other way around. First put $3$ birds, that is the maximum amount without being overpopulated into the cages. This results in $3cdot 21=63$ birds placed. At this point no cages are overpopulated but we still have $100-63=37$ birds to place. From the text it seems that it doesn´t matter wether we overpopulate by one or five birds per cage (which is definitely not like in real life). So I would start filling the cages up to $10$ with the remaining birds.

Can you figure out the rest?

You have $37$ more birds to place and each cage can hold $7$ more birds, since $37=5cdot 7+2$ you can fill up $5$ cages to full and put the last $2$ birds in a $6$th cage ending up with $6$ overpopulated cages

share|cite|improve this answer

edited 36 mins ago

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

2,8421030

$endgroup$

add a comment | 

0

$begingroup$

I would think the other way around. First put $3$ birds, that is the maximum amount without being overpopulated into the cages. This results in $3cdot 21=63$ birds placed. At this point no cages are overpopulated but we still have $100-63=37$ birds to place. From the text it seems that it doesn´t matter wether we overpopulate by one or five birds per cage (which is definitely not like in real life). So I would start filling the cages up to $10$ with the remaining birds.

Can you figure out the rest?

You have $37$ more birds to place and each cage can hold $7$ more birds, since $37=5cdot 7+2$ you can fill up $5$ cages to full and put the last $2$ birds in a $6$th cage ending up with $6$ overpopulated cages

share|cite|improve this answer

edited 36 mins ago

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

2,8421030

$endgroup$

add a comment | 

$begingroup$

I would think the other way around. First put $3$ birds, that is the maximum amount without being overpopulated into the cages. This results in $3cdot 21=63$ birds placed. At this point no cages are overpopulated but we still have $100-63=37$ birds to place. From the text it seems that it doesn´t matter wether we overpopulate by one or five birds per cage (which is definitely not like in real life). So I would start filling the cages up to $10$ with the remaining birds.

Can you figure out the rest?

You have $37$ more birds to place and each cage can hold $7$ more birds, since $37=5cdot 7+2$ you can fill up $5$ cages to full and put the last $2$ birds in a $6$th cage ending up with $6$ overpopulated cages

share|cite|improve this answer

edited 36 mins ago

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

2,8421030

$endgroup$

share|cite|improve this answer

edited 36 mins ago

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

2,8421030

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

2,8421030

answered 1 hour ago

Vinyl_coat_jawaVinyl_coat_jawa

2,8421030

0

$begingroup$

Fill all $21$ cages with $3$ birds.

That leaves $100 - 3cdot 21 = 37$ birds loose.

Cram $7$ birds into each of $37/7 = 5 + 2/7$ cages.

Thus $5cdot 10 + 5 + 1cdot 3 = 100$ gives $6$ overpopulated cages.

How you filled the cages does not total to $100$.

share|cite|improve this answer

edited 35 mins ago

Vinyl_coat_jawa

2,8421030

answered 55 mins ago

William ElliotWilliam Elliot

8,2872720

$endgroup$

add a comment | 

0

$begingroup$

Fill all $21$ cages with $3$ birds.

That leaves $100 - 3cdot 21 = 37$ birds loose.

Cram $7$ birds into each of $37/7 = 5 + 2/7$ cages.

Thus $5cdot 10 + 5 + 1cdot 3 = 100$ gives $6$ overpopulated cages.

How you filled the cages does not total to $100$.

share|cite|improve this answer

edited 35 mins ago

Vinyl_coat_jawa

2,8421030

answered 55 mins ago

William ElliotWilliam Elliot

8,2872720

$endgroup$

add a comment | 

$begingroup$

Fill all $21$ cages with $3$ birds.

That leaves $100 - 3cdot 21 = 37$ birds loose.

Cram $7$ birds into each of $37/7 = 5 + 2/7$ cages.

Thus $5cdot 10 + 5 + 1cdot 3 = 100$ gives $6$ overpopulated cages.

How you filled the cages does not total to $100$.

share|cite|improve this answer

edited 35 mins ago

Vinyl_coat_jawa

2,8421030

answered 55 mins ago

William ElliotWilliam Elliot

8,2872720

$endgroup$

share|cite|improve this answer

edited 35 mins ago

Vinyl_coat_jawa

2,8421030

answered 55 mins ago

William ElliotWilliam Elliot

8,2872720

edited 35 mins ago

Vinyl_coat_jawa

2,8421030

edited 35 mins ago

Vinyl_coat_jawa

2,8421030

answered 55 mins ago

William ElliotWilliam Elliot

8,2872720

answered 55 mins ago

William ElliotWilliam Elliot

8,2872720