20 stallions, 0 mares and 80 ponies.
17 stallion, 5 mares and 78 ponies.
Stallion 14, mare 10 and pony 76.
Stallion 1 1, mare 15, pony 74.
8 stallions, 20 mares and 72 ponies.
Five stallions, 25 mares and 70 ponies.
2 stallions, 30 mares and 68 ponies.
The reasoning process is as follows: (The method may be irregular)
Suppose the stallion is X, the mare is Y and the pony is Z.
Two equations are obtained: 3X+2Y+Z/2= 100 (1).
X+Y+Z= 100 (2)
From (2): Z= 100-X-Y (3)
Substituting (3) into (2) gives 3x+2y+(100-x-y)/2 =100 (4).
Simplify equation (4): 5X+3Y= 100.
Because horses are whole, x and y must both be integers.
It can be concluded that 3Y must be a multiple of 5, and only 15, 30, 45, 60, 75, 90 can have a common multiple of 3 and 5 below 100.
Suppose 3Y= 15, then Y=5, X= 17, then Z=78.
Suppose 3Y=30, then Y= 10, X= 14, then Z=76.
Suppose 3Y=45, then Y= 15, X= 1 1, then Z=74.
Suppose 3Y=60, then Y=20, X=8, then Z=72.
Suppose 3Y=75, then Y=25, X=5, then Z=70.
Suppose 3Y=90, then Y=30, X=2, then Z=68.
Another possibility is that 100 can be divisible by 5X when 3Y=0.
At this time, Y=0 and X=20, so Z=80.