Question 281733
you know there are going to be a lot of chickens and few cows.
w+p+n=100
10w+3p+.5n=100

If there were all chickens he could buy 200 chickens
If there were all cows he could by 10 cows
If there were all pigs he could by 33 pigs and have a dollar left over.

So the numbers need to be between 0 and those upper limits.
0 < w < 11
0 < p < 34
0 < n < 201
chicken=n=94
cow=w=5
pig=p=1
5*10+94*.5+1*3=100
50+47+3=100
94+5+1=100