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# Tag Archives: Chinese Puzzle

## The problem of the Hundred Fowls

This is a popular Chinese problem, on Linear Diophantine equations, which in wording seems as a puzzle or riddle. However, when used algebraic notations, it looks obvious. The problems states :

 If a cock is worth 5 coins, a hen 3 coins, and three chickens together 1 coin, how many cocks, hens and chickens, totaling 100 in number, can be bought for 100 coins?

This puzzle in terms of algebraic equations can be written as $5x+3y+\frac{1}{3}z=100$ and $x+y+z=100$
where $x, y, z$ being the number of cocks, hens and chicks respectively.
We find that there are two equations with three unknown quantities. So eliminating one of the unknowns, by putting $z=100-x-y$ from second equation into first one such that $5x+3y+\frac{1}{3} (100-x-y)=100$
or, $15x+9y+100-x-y=300$
or, $14x+8y=200$
or, $7x+4y=100$.
Which is a linear Diophantine equation (with only two unknown quantities).
The equation $7x+4y=100$ has the general solution   [links to WolframAlpha] $x=4 t$ and $y=25-7t$, so that $z=75+3t$ where $t$ is an arbitrary integer.
Now, since $x, y, z$ are the number of creatures, hence $x, y, z >0$ and thus $4t >0$ , $25-7t >0$ and $75+3t >0$ which imply that $0 < t < 3\frac{4}{7}$. And because t must have integer values, we have $t=1,2,3$. Which gives the following three solutions:

 Values of $t$ No. Of cocks ( $x=4 t$ ) No. Of hens ($y=25-7t$) No. Of chicks ($z=75+3t$) 1 4 18 78 2 8 11 81 3 12 4 84

So there are the three ways to chose the number of cocks, hens and chicken totaling 100 to buy for 100 coins.

Problem Sources:
Elementary Number Theory
David M. Burton, 2006
McGrawHill Publications

Wikipedia article on Diophantine Equations

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