$100$ birds in $21$ cages each with $≤ 10$, with least cages having $≥ 4$ birds?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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
edited Feb 25 at 6:51
user21820
39.7k543157
asked Feb 24 at 11:09
GodlixeGodlixe
1157
$endgroup$
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
edited Feb 25 at 6:51
user21820
39.7k543157
asked Feb 24 at 11:09
GodlixeGodlixe
1157
$endgroup$
5
$begingroup$
You put only $98$ birds...
$endgroup$
– user289143
Feb 24 at 11:15
2
$begingroup$
The way you think here is correct altough you missed $2$ birds, adding them into the next cage gives the correct answer
$endgroup$
– Vinyl_cape_jawa
Feb 24 at 12:01
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
edited Feb 25 at 6:51
user21820
39.7k543157
asked Feb 24 at 11:09
GodlixeGodlixe
1157
$endgroup$
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
combinatorics discrete-mathematics pigeonhole-principle discrete-optimization
edited Feb 25 at 6:51
user21820
39.7k543157
asked Feb 24 at 11:09
GodlixeGodlixe
1157
edited Feb 25 at 6:51
user21820
39.7k543157
asked Feb 24 at 11:09
GodlixeGodlixe
1157
edited Feb 25 at 6:51
user21820
39.7k543157
edited Feb 25 at 6:51
user21820
39.7k543157
edited Feb 25 at 6:51
user21820
39.7k543157
39.7k543157
asked Feb 24 at 11:09
GodlixeGodlixe
1157
asked Feb 24 at 11:09
GodlixeGodlixe
1157
asked Feb 24 at 11:09
GodlixeGodlixe
1157
1157
5
$begingroup$
You put only $98$ birds...
$endgroup$
– user289143
Feb 24 at 11:15
2
$begingroup$
The way you think here is correct altough you missed $2$ birds, adding them into the next cage gives the correct answer
$endgroup$
– Vinyl_cape_jawa
Feb 24 at 12:01
add a comment |
5
$begingroup$
You put only $98$ birds...
$endgroup$
– user289143
Feb 24 at 11:15
2
$begingroup$
The way you think here is correct altough you missed $2$ birds, adding them into the next cage gives the correct answer
$endgroup$
– Vinyl_cape_jawa
Feb 24 at 12:01
5
5
$begingroup$
You put only $98$ birds...
$endgroup$
– user289143
Feb 24 at 11:15
$begingroup$
You put only $98$ birds...
$endgroup$
– user289143
Feb 24 at 11:15
2
2
$begingroup$
The way you think here is correct altough you missed $2$ birds, adding them into the next cage gives the correct answer
$endgroup$
– Vinyl_cape_jawa
Feb 24 at 12:01
$begingroup$
The way you think here is correct altough you missed $2$ birds, adding them into the next cage gives the correct answer
$endgroup$
– Vinyl_cape_jawa
Feb 24 at 12:01
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Say we have $n$ cages each with $3$ birds, then we must fill the other $21-n$ cages 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 $6$ overpopulated cages.
answered Feb 24 at 11:21
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
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 $5 cdot 7 < 37 < 6 cdot 7$)
answered Feb 24 at 11:19
user289143user289143
1,000313
$endgroup$
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.
answered Feb 24 at 11:17
PeterPeter
48.9k1239136
$endgroup$
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
answered Feb 24 at 11:20
Vinyl_cape_jawaVinyl_cape_jawa
3,33011433
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Say we have $n$ cages each with $3$ birds, then we must fill the other $21-n$ cages 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 $6$ overpopulated cages.
answered Feb 24 at 11:21
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
add a comment |
$begingroup$
Say we have $n$ cages each with $3$ birds, then we must fill the other $21-n$ cages 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 $6$ overpopulated cages.
answered Feb 24 at 11:21
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
add a comment |
$begingroup$
Say we have $n$ cages each with $3$ birds, then we must fill the other $21-n$ cages 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 $6$ overpopulated cages.
answered Feb 24 at 11:21
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
Say we have $n$ cages each with $3$ birds, then we must fill the other $21-n$ cages 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 $6$ overpopulated cages.
answered Feb 24 at 11:21
Maria MazurMaria Mazur
47.9k1260120
edited Feb 24 at 15:27
answered Feb 24 at 11:21
Maria MazurMaria Mazur
47.9k1260120
answered Feb 24 at 11:21
Maria MazurMaria Mazur
47.9k1260120
answered Feb 24 at 11:21
Maria MazurMaria Mazur
47.9k1260120
47.9k1260120
add a comment |
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 $5 cdot 7 < 37 < 6 cdot 7$)
answered Feb 24 at 11:19
user289143user289143
1,000313
$endgroup$
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 $5 cdot 7 < 37 < 6 cdot 7$)
answered Feb 24 at 11:19
user289143user289143
1,000313
$endgroup$
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 $5 cdot 7 < 37 < 6 cdot 7$)
answered Feb 24 at 11:19
user289143user289143
1,000313
$endgroup$
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 $5 cdot 7 < 37 < 6 cdot 7$)
answered Feb 24 at 11:19
user289143user289143
1,000313
answered Feb 24 at 11:19
user289143user289143
1,000313
answered Feb 24 at 11:19
user289143user289143
1,000313
answered Feb 24 at 11:19
user289143user289143
1,000313
1,000313
add a comment |
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.
answered Feb 24 at 11:17
PeterPeter
48.9k1239136
$endgroup$
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.
answered Feb 24 at 11:17
PeterPeter
48.9k1239136
$endgroup$
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.
answered Feb 24 at 11:17
PeterPeter
48.9k1239136
$endgroup$
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.
answered Feb 24 at 11:17
PeterPeter
48.9k1239136
answered Feb 24 at 11:17
PeterPeter
48.9k1239136
answered Feb 24 at 11:17
PeterPeter
48.9k1239136
answered Feb 24 at 11:17
PeterPeter
48.9k1239136
48.9k1239136
add a comment |
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
answered Feb 24 at 11:20
Vinyl_cape_jawaVinyl_cape_jawa
3,33011433
$endgroup$
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
answered Feb 24 at 11:20
Vinyl_cape_jawaVinyl_cape_jawa
3,33011433
$endgroup$
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
answered Feb 24 at 11:20
Vinyl_cape_jawaVinyl_cape_jawa
3,33011433
$endgroup$
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
answered Feb 24 at 11:20
Vinyl_cape_jawaVinyl_cape_jawa
3,33011433
edited Feb 24 at 11:45
answered Feb 24 at 11:20
Vinyl_cape_jawaVinyl_cape_jawa
3,33011433
answered Feb 24 at 11:20
Vinyl_cape_jawaVinyl_cape_jawa
3,33011433
answered Feb 24 at 11:20
Vinyl_cape_jawaVinyl_cape_jawa
3,33011433
3,33011433
add a comment |
add a comment |
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5
$begingroup$
You put only $98$ birds...
$endgroup$
– user289143
Feb 24 at 11:15
2
$begingroup$
The way you think here is correct altough you missed $2$ birds, adding them into the next cage gives the correct answer
$endgroup$
– Vinyl_cape_jawa
Feb 24 at 12:01