(related to Problem: Judkins's Cattle)
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 from 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.
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