TITLE: $100$ birds in $21$ cages each with $≤ 10$, with least cages having $≥ 4$ birds?
QUESTION [9 upvotes]: 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!

REPLY [11 votes]: 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.