Example 5 At the end of the fifth century, the ancient Chinese mathematician Zhang Qiujian proposed an indeterminate equation problem in his compiled "Suanjing" — a famous "Hundred Chickens Problem" in the history of world mathematics:

A rooster is worth five coins, a hen is worth three coins, and three chicks are worth one coin. With a hundred coins to buy a hundred chickens, how many roosters, hens, and chicks should be bought? (Translation: The value of each rooster is 5 coins, the value of each hen is 3 coins, and the value of every three chicks is 1 coin. With 100 coins to buy 100 chickens, how many roosters, hens, and chicks should be bought?)
Let $x, y, z$ represent the number of roosters, hens, and chicks, respectively. We then have the following equations:
$$\begin{array}{l}
5 x+3 y+\frac{z}{3}=100 \\
x+y+z=100
\end{array}$$

From (16) and (17), we have
$$15 x+9 y+z-x-y-z=14 x+8 y=200$$

That is,
$$7 x+4 y=100$$

To solve this problem, we need to find the non-negative integer solutions to the above equation. First, solve
$$7 s+4 t=1$$

Since $7=4+3, 4=3+1$, we get
$$1=4-3=4-(7-4)=-7+4 \times 2$$

Therefore, a solution to (19) is $s=-1, t=2$. Let $x=100 s$, $y=100 t$, then from (19) we have
$$7 x+4 y=100(7 s+4 t)=100$$

Hence, a solution to (18) is $x=-100, y=200$. By Theorem 1, all integer solutions to (18) can be expressed as
$$\begin{array}{c}
x=-100-4 t, \quad y=200+7 t \\
(t=0, \pm 1, \pm 2, \cdots)
\end{array}$$

Since $x$ and $y$ represent the number of roosters and hens, we must have $x \geqslant 0, y \geqslant 0$. From $-100-4 t \geqslant 0$ we get $4 t \leqslant-100$, and from $200+7 t \geqslant 0$ we get $7 t \geqslant-200$, thus
$$-\frac{200}{7} \leqslant t \leqslant-25$$

Since $t$ is an integer, $t=-28,-27,-26,-25$. The number of chicks is given by (17) as
$$\begin{aligned}
z= & 100-x-y=100-(-100-4 t)-(200 \\
& +7 t)=-3 t
\end{aligned}$$

From (20), for $t=-28,-27,-26,-25$ and $z=-3 t$, we obtain the following four sets of solutions:
$$\left.\left.\left.\left.\begin{array}{l}
x=12 \\
y=4 \\
z=84
\end{array}\right\}, \begin{array}{l}
x=8 \\
y=11 \\
z=81
\end{array}\right\}, \begin{array}{l}
x=4 \\
y=18 \\
z=78
\end{array}\right\}, \begin{array}{l}
x=0 \\
y=25 \\
z=75
\end{array}\right\}$$