5x+3y+( 100-x-y)/3 = 100
7x+4y after completion =100
Because both x and y must be integers, and 4y and 100 are multiples of 4, 7x must also be multiples of 4.
Therefore, there are several groups of solutions that meet the requirements initially: x = 4, y =18; Or x = 8, y =11; Or x = 12, y = 4.
In addition, it must be satisfied that the amount of money for buying chickens is also an integer, that is, (100-x-y)/3 is an integer, and all of them meet the requirements when substituted into the first three groups.
Therefore, there are three solutions to this problem:
(1) rooster 4, hen 18, chicken 78
(2) Rooster 8, hen 1 1, chicken 8 1
(3) rooster 12, hen 4, chick 84