To solve this problem we need to start by turning
the available information into equations. Let the number of animals be *c*,
*p*, *h*, and *r* for cows, pigs, chickens, and rabbits respectively.
Then we have:

c+p+h+r= 100

10c+ 3p+h+ 0.5r= 100

c>p

h>r

Let's also assume that there is at least 1 of each
animal, since otherwise it would not be on the list. From the second equation we can see that 1 <= *c* <= 9 (i. e.,
the number of cows must be between 1 and 9). In fact, since *c* > *p*
and *p* > 0, there must be at least 2 cows. This leads to 8 cases as follows:

Case 1: *c* = 2

2 +p+h+r= 100

20 + 3p+h+ 0.5r= 100

Therefore, –2p+ 0.5r= 18

Since the number of pigs is less than the number of cows,p= 1. Sor= 40, andh= 57. This solution satisfies all the conditions of the problem. But is it unique?

Case 2: *c* = 3

3 +p+h+r= 100

30 + 3p+h+ 0.5r= 100

Therefore, –2p+ 0.5r= 27

Solving forrgivesr= 54 + 4p. Sincep> 0, the number of rabbits must be greater than 54. But ifr> 54, then there cannot be more chickens than rabbits because that would put the total number of animals over 100. Therefore, Case 2 fails to give a solution.

Cases 3–8: *c* > 3

Combining the first two equations wherecis known gives

–2p+ 0.5r= (100 –c) – (100 – 10c) = 9c

Solving forrgivesr= 18c+ 4p. Ascincreases,rincreases too, so we can apply the reasoning from Case 2 to show thatris never less than 50, sohcannot be greater thanr. Therefore, none of the remaining cases lead to a solution.

The solution from Case 1 is unique. The list must have called for:

- 2 Cows @
- $10 ea. = $20
- 1 Pig @
- $3 ea. = $3
- 57 Chickens @
- $1 ea. = $57
- 40 Rabbits @
- 50¢ ea. = $20

100 animals for $100