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In a well known puzzle, a father willed to his three sons $17$ camels, with the proviso that $\frac{1}{2}$ of the inheritance should go to the oldest among them, with $\frac{1}{3}$ being due to the middle one and $\frac{1}{9}$ to the youngest. Shortly after the father's death, a wise man riding on his camel through the village noticed the three brothers in a quandary. He added his camel to the inherited $17,$ thus getting a herd of $18$ animals. He gave $\frac{1}{2}$ of these (i. e. $9$ camels) to the oldest brother, $6$ $(= \frac{18}{3})$ to the middle one, and $2$ $(= \frac{18}{9})$ to the youngest. $1$ camel remained $(1 = 18 - 9 - 6 - 2),$ which he climbed up and rode away on. To the great satisfaction of all brothers, each of them received more than was willed by their father.

Among those who found interest in the puzzle, one may try solving it before reading the solution; another may read and enjoy the solution; somebody else may think up related problems. Paul Stockmeyer - a professor emeritus of computer science at the College of William and Mary - did just that (Math Horizons, Volume 21, Number 1, September 2013 , pp. 8-11), posing a generalized question and even listing several related research problems.

Are there situations that include two brothers? other problems with three? problems with four, etc. How many camels could be shared in such a manner? Below are several identities that answer some of the questions, if only partially. Rather obviously the problem involves the unit fractions:

Two sons, three camels: $\frac{1}{2} + \frac{1}{4} = 1 - \frac{1}{4}$

Two sons, five camels: $\frac{1}{2} + \frac{1}{3} = 1 - \frac{1}{6}$

Three sons, seven camels: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = 1 - \frac{1}{8}$

Three sons, eleven camels: $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} = 1 - \frac{1}{12}$

Three sons, eleven camels: $\frac{1}{2} + \frac{1}{3} + \frac{1}{12} = 1 - \frac{1}{12}$

Three sons, nineteen camels: $\frac{1}{2} + \frac{1}{4} + \frac{1}{5} = 1 - \frac{1}{20}$

Four sons, 1805 camels: $\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} = 1 - \frac{1}{1806}$

There are many more examples. But could an estate consist of 9 or 21 or of an even number of camels. Are there divisions where the oldest son is willed one third of the inheritance and not one half? What about one fourth?

This is amazing how many questions may be asked starting with such an elementary problem. Probably not all problem submit to such extensions, but many do. I've been recently referred to the site of an Online (more accurately, by correspondence) Russian Math Olympiad honoring Leonhard Euler. Here are some examples from this year's olympiad:

1. In the expression $\displaystyle\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\cdots *\frac{98}{99}*\frac{99}{100}$ replace each star with one of four arithmetic operations to make the result zero.

   What comes to mind? First, an obvious generalization (that gives away a solution) is to end the product with $\displaystyle\frac{n^2-1}{n^2}.$ Then it's natural to ask if there are other cases. I am not sure of the answer yet. What I am sure about is that a solution - if exists - is never unique. Prove that.

2. In the island of knights (that always tell truth) and knaves (that always lie), $22$ fellows stand in a circle, all facing the center, and each declares that the ten to his left are all knaves. How many knaves are actually there?

   Is the relation $1+10+1+10=22$ inherent for the solvability of the problem?

3. Draw five staright lines in a plane that intersect in exactly seven points.

   What number of intersections is possible for $n$ lines?