



 
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  


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