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
could be made up in one of two ways--1 ox, 23 pigs, and 96 sheep, or 3
oxen, 8 pigs, and 109 sheep. But the first is excluded by the statement
that the animals consisted of "oxen, pigs, and sheep," because a single
ox is not oxen. Therefore the second grouping is the correct answer.