www.TheMathWebSite.com       Enrichment Problems HD
      Answer is at Bottom of Page.


      Cattle Dealer :

Hiram 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?



















Copyright © 1994-. All rights reserved.    www.TheMathWebSite.com       HD2


----------------- Cut or Fold Page Back Along This Line -----------------------

      Cattle Dealer :     ANSWER

As there were five droves with an equal number of animals in each drove,
the number must be divisible by 5; and as every one of the eight dealers
bought the same number of animals, the number must be divisible by 8.
Therefore the number must be a multiple of 40. The highest possible
multiple of 40 that will work will be found to be 120, and this number,
with more than 1 ox, must be 3 oxen, 8 pigs, and 109 sheep.