The Ranchers’ Split

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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Next Up

What’s the next number in this sequence?

1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, …

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Custom Baking

From a Russian puzzle collection:

Is it possible to bake a cake that can be divided into four parts by a single straight cut?

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Black and White

orbán chess puzzle

Tibor Orbán offered this puzzle in Die Schwalbe in 1976. The position above can be reached in exactly 3 moves in several ways — for example:

1. e4 c6 2. Bb5 e6 3. Bxc6 dxc6
1. e4 e6 2. Bc4 c6 3. Bxe6 dxe6

How can it be reached in exactly 4 moves?

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Trip Planning

mouse puzzle

A mouse wants to eat his way through a 3 × 3 × 3 cube of cheese, starting in one of the corners and tunneling through all 27 1 × 1 × 1 sub-cubes, visiting each once. Can he arrange his route so that he finishes at the center of the cube? Assume that he always moves between orthogonally adjacent cubes, traveling through walls but not through edges or corners.

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Containing an Arc

arc puzzle

University of Illinois mathematician John Wetzel called this one of his favorite problems in geometry. Call a plane arc special if it has length 1 and lies on one side of the line through its end points. Prove that any special arc can be contained in an isosceles right triangle of hypotenuse 1.

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A Triangle Puzzle

posamentier puzzle

In isosceles triangle ABC, CD = AB and BE is perpendicular to AC. Show that CEB is a 3-4-5 right triangle.

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Black and White

neuhaus chess problem

By Eugene Neuhaus Jr. White to mate in two moves.

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