$100$ birds in $21$ cages each with $≤ 10$, with least cages having $≥ 4$ birds? 
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!
 A: 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 $5 \cdot 7 < 37 < 6 \cdot 7$)
A: 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. 
A: 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 $3\cdot 21=63$ 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 $37=5\cdot 7+2$ 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

A: Say we have $n$ cages each with $3$ birds, then we must fill the other $21-n$ cages with $100-3n$ birds. So $$100-3n\leq (21-n)\cdot 10\implies n\leq 15$$
$n= 15$ is achiavable: $15\cdot 3+5\cdot 10+1\cdot 5 =100$. 
So you must have at least $6$ overpopulated cages.
