A problem from the Graham Dial, a publication of Graham Transmissions Inc., via L.A. Graham’s The Surprise Attack in Mathematical Problems (1968):
Little Euclid went with his father to visit a friend who ran a small ranch. The rancher told them that he and his partner used to herd cattle but had then sold the herd for as many dollars per head as there were head in the herd. With the proceeds they had bought as many sheep as they could at $10.00 per sheep, and with the money that remained they bought a dog. Later the rancher and his partner decided to split up evenly, so they added the dog to the herd and each of them took the same number of animals.
Euclid surprised him by saying, “I hope you paid him $2.00 to even things up.”
“Why, yes, I did,” the rancher said, “but how did you know? I haven’t told you any of the numbers or prices.” Can you prove that Little Euclid had to be correct?
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First, three observations: (1) The number of dollars that the men received for the herd of cattle was a perfect square. (2) The number of sheep they bought was odd, because it was necessary to throw in the dog in order to divide the animals evenly. (3) Since they bought an odd number of sheep at $10.00 per head, the next-to-last digit in the square payment from observation (1) must be odd.
Now, as it happens, in any perfect square, if the next-to-last digit is odd, the last figure must be a 6. The simplest way to see this is to consider that any number can be expressed as 10a + b, where a comprises all the digits except the last and b is the last digit (0 to 9). The square of this number is 100a2 + 20ab + b2. But 100a2 has zeros in the last two places, and 20ab has a zero in the last place and is preceded by an even number. So, in order for the square to have an odd digit in the next-to-last position, b2 must have an odd first digit, and the only possibilities are 42 = 16 and 62 = 36.
In either case, the dog cost $6.00, and the rancher owed his partner $2.00 for accepting a $6.00 dog in place of a $10.00 sheep.
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