Thursday, August 12, 2010

Thursday Puzzle: Judkins's Cattle

Another one from Dudeney:


"Hiram B. Judkins, a cattle-dealer of Texas, had five droves of animals, consisting of oxen, pigs, and sheep, with the same number of animals in each drove. One morning he sold all that he had to eight dealers. Each dealer bought the same number of animals, paying seventeen dollars for each ox, four dollars for each pig, and two dollars for each sheep; and Hiram received in all three hundred and one dollars. What is the greatest number of animals he could have had? And how many would there be of each kind?"

2 comments:

Joshua Fahey said...

1 ox
23 pigs
96 sheep

for 120 animals as the greatest possible number

Eye of the Frog said...

Unless it's required to have one of each animal in each drove (I couldn't tell from the question), then we have:

5 oxen
33 pigs
42 sheep

for 80 animals.

BTW, nice picture of Shaun.

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