Hi Jessica,

Solving this problem involves some guess and check but a little algebra can reduce the amount of guessing and checking.

Suppose the number of chicks purchased is $C,$ the number of pigs purchased is $p$ and the number of sheep purchased is $s.$ Since the total number of animals purchased is 100 you know that

\[c + p + s = 100.\]

You also know that the total cost was $\$100$ and hence

\[0.1 c + 2 p + 5 s = 100.\]

Multiplying the second equation by $10$ and subtracting the first equation from the result yields

\[19 p + 49 s = 900\]

or

\[19 p = 900 - 49s.\]

$s$ and $p$ are positive integers and $s$ can't be larger then 18. Do you see why? Now comes the guess and check. Try possible values of $s$ until you find an $s$ so that $900 - 49s$ is a multiple of 19. Does this value of $s$ give you a solution to the problem?

Penny