If a cock is worth $5$ Rs., a hen $3$ Rs., and three chicks together $1$ Rs. How many cocks, hens and chicks totaling $100$ can be bought for $100$ Rs.

Example:  If a cock is worth $5$ Rs., a hen $3$ Rs., and three chicks together $1$ Rs. How many cocks, hens and chicks totaling $100$ can be bought for $100$ Rs.

Proof:  Suppose that $x$ is the number of cocks, $y$ is the number of hen and $z$ is the number of chicks totaling $100$ bought for $100$ Rs. Therefore we will get following equations:

$x + y + z = 100$ and  $5x +3y + \dfrac{1}{3} z = 100$

We put $z = 100 - x - y$ in $5x +3y + \dfrac{1}{3} z = 100$.

So $5x + 3y + \dfrac{1}{3} (100 - x - y) = 100$

Multiply both side by $3$

$15x +9y +100 - x -y = 300$

$14x +8y = 200$

$7x + 4y = 100$. So we got the linear equation $7x + 4y = 100$. Since $gcd(7,4) = 1$ divides $100$, linear equation $7x + 4y = 100$ is a Diophantine equation and it has a solution.

General solution of the equation $7x + 4y = 100$ is $x =4t, ~ y = 25 - 7t$ and hence $z= 75 + 3t$, where $t$ is an arbitrary integer. 

Therefore $x = 4,~y = 18 $ and $z = 78$ is a one solution of given problem.  Therefore $4$ is the number of cocks, $18$ is the number of hen and $78$ is the number of chicks totaling $100$ bought for $100$ Rs. There are other solutions $x = 5, ~y = 11$ and $z = 81$, another one $x = 12, ~ y = 4, ~ z = 84$.

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