{ The Monkey and the Coconuts Martin Gardner said that the Monkey and the Coconuts is "probably the most worked on and least often solved" algebraic puzzle. According to the OMNI magazine (April 1991), this puzzle caused quite a commotion when it appeared in the October 9, 1926 Saturday Evening Post as part of a short story, "Coconuts", by Ben Ames Williams. Williams ended his story without solving the puzzle. It is said that some 2000 letters came to the Saturday Evening Post offices the week after the "Coconuts" appeared, all requesting the puzzle solution. Here is the puzzle itself: Five sailors are shipwrecked on a desert island. They quickly determine that the only other inhabitant of the island is a monkey and that the only food is coconuts. They set about collecting as many coconuts as they can and put them all in a pile. By nightfall they are too tired to divide the harvest; so they agree to go to sleep and divvy up the coconuts the next morning. During the night one sailor awakens, suspicious that the others might try to cheat him, and desides to take his portion then and there and not wait until morning. He divides the coconuts into five piles and finds there is one coconut left over, which he gives to the monkey. He hides one of the five piles, then puts the rest of the nuts together and returns to sleep. About an hour later a second sailor awakens with the same suspicions and does the same thing: He divides the coconuts into five piles, leaving one extra, which he gives to the monkey. Then he hides what he thinks is his share and goes back to sleep. One after another the rest of the sailors do the same: they each take one fifth of the coconuts in the pile (after giving the extra one to the monkey) and then return to sleep. When the sailors awaren the next morning they all notice the coconut pile is much smaller than it was the night before, but since each man is as guilty as the others, no one says anything. They divide the coconuts (for the sixth time) and again there is one left for the monkey. The question is: How many coconuts were in the original pile? The program below solves the puzzle by asking the query all SailorsMonkeyCoconuts(coconuts) Note that there is an infinite number of solutions, the program finds all solutions smaller than specified by the type declaration of Coconuts. (The smallest positive number that solves the puzzle is 15621.) } Coconuts = L[0..50000] // Extend the range for more solutions pred SailorsMonkeyCoconuts(pile0::Coconuts) iff // first sailor: divide the pile of the coconuts (pile0) into 5 parts // (p1 each), give 1 coconut to the monkey, hide 1/5 of the coconuts, // leaving the resulting pile (pile1) 4 parts of coconuts (4*p1) p1::Coconuts & pile0 = 5*p1 + 1 & pile1 = 4*p1 & // second sailor: divide the pile of the coconuts (pile1) into 5 parts // (p2 each), give 1 coconut to the monkey, hide 1/5 of the coconuts, // leaving the resulting pile (pile2) 4 parts of coconuts (4*p2) p2::Coconuts & pile1 = 5*p2 + 1 & pile2 = 4*p2 & // third sailor: divide the pile of the coconuts (pile2) into 5 parts // (p3 each), give 1 coconut to the monkey, hide 1/5 of the coconuts, // leaving the resulting pile (pile3) 4 parts of coconuts (4*p3) p3::Coconuts & pile2 = 5*p3 + 1 & pile3 = 4*p3 & // fourth sailor: divide the pile of the coconuts (pile3) into 5 parts // (p4 each), give 1 coconut to the monkey, hide 1/5 of the coconuts, // leaving the resulting pile (pile4) 4 parts of coconuts (4*p4) p4::Coconuts & pile3 = 5*p4 + 1 & pile4 = 4*p4 & // fifth sailor: divide the pile of the coconuts (pile4) into 5 parts // (p5 each), give 1 coconut to the monkey, hide 1/5 of the coconuts, // leaving the resulting pile (pile5) 4 parts of coconuts (4*p5) p5::Coconuts & pile4 = 5*p5 + 1 & pile5 = 4*p5 & // final split: divide the final pile of the coconuts (pile5) into 5 parts // (p6 each), give 1 coconut to the monkey p6::Coconuts & pile5 = 5*p6 + 1

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