5 equal droves, 8 equal buyers, $301 total — how many oxen, pigs and sheep?
Texas cattle-dealer Hiram Judkins had five droves of oxen, pigs and sheep, each drove the exact same size. One morning he sold the lot to eight dealers, each buying an equal number of animals, for $17 per ox, $4 per pig and $2 per sheep — $301 in all. What's the greatest number of animals he could have had, and how many of each kind?
Reveal the answer
120 animals: 3 oxen, 8 pigs and 109 sheep (3×$17 + 8×$4 + 109×$2 = $51+$32+$218 = $301; 120 splits evenly into 5 droves of 24 and 8 sales of 15). A second combination fits the arithmetic too, but needs only 1 ox — and Dudeney ruled 'one' isn't 'oxen.' From Henry Dudeney's Amusements in Mathematics (1917), puzzle No. 35.
— Henry Dudeney, Amusements in Mathematics — Thomas Nelson & Sons, 1917, Puzzle No. 35
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