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My friend who owns a farm near Dallas had five droves of animals on his farm consisting of cows, sheep and pigs with the same number of animals in each drove. One day he sold them all to eight dealers. Each of the eight dealers bought the same number of animals and paid at the rate of 17 Dollars for each cow, 2 Dollars for each sheep, and 2 Dollars for each pig. The money he received in total was 285 Dollars.

How many animals did he own, and how many of each kind?

Edited for proper punctuation.

Edited by brhan
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since the animals need to be able to be split into 5 groups or 8 equal groups, the number of animals needs to be a multiple of 40.

If there are 40 animals, there would need to be 13 2/3 cows to get 285 bucks.....nope.

If there are 80 animals, there would need to be 8 1/3 cows to get 285 bucks.....still not a whole number.

If there are 120 animals, there would need to be 3 cows. It worked!

If there are 160 animals, there would need to be -2 1/3. Hmm . ..negative cows.

So there must be 120 total animals, 3 of which are cows. the other 117 are pigs and sheep but there isn't enough information (since their cost is the same) to determine the exact amounts of those.

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3 cows, x sheep, (117-x) pigs

First since the total number of animals is evenly divisible by both 5 and 8, it's obviously a multiple of 40. Also, since the total cost of the animals is odd, the number of cows must also be odd. A minute of trial and error results yields that there are 3 cows ($51). Since sheep and pigs are both priced at $2, it is impossible to distinguish them apart from this problem. However, there are 117 pigs and sheep combined ($234). This gives us 120 animals total that came in droves of 24 and sold in groups of 15.

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