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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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