16. $A, B$ together hunted 10 birds, the product of the squares of the number of bullets used by the two is 2880, and the product of the number of bullets used is 48 times the product of the number of birds they hunted. If the number of bullets used by the two were exchanged, then $B$ would get 5 more than $A$, find how many birds each got as follows?
Let $x, y$ represent the number of bullets used by $A, B$ respectively. Let $A$ use $u$ bullets to hit 1 bird, and $B$ use $v$ bullets to hit 1 bird. Then $\frac{x}{u}$ and $\frac{y}{v}$ represent the number of birds obtained. Thus, we have
$$
\begin{array}{l}
x^{2}+y^{2}=2880, \\
x y=\frac{48 x y}{u v}, \text { i.e., } u v=48 .
\end{array}
$$

According to the problem, $\frac{x}{u}+\frac{y}{v}=10, \frac{x}{v}-\frac{y}{u}=5$.
From the last two equations, we get $\frac{x^{2}+y^{2}}{u}=10 x-5 y$,
i.e., $u(2 x-y)=576$,
$$
\frac{x^{2}+y^{2}}{v}=10 y+5 x,
$$

i.e., $v(2 y+x)=576$.
Therefore, $u v(2 x-y)(2 y+x)=576 \times 576$.
$$
\text { Hence, }(2 x-y)(2 y+x)=12 \times 576 \text {. }
$$

Dividing both sides by $x^{2}+y^{2}=2880$ gives
$$
\frac{(2 x-y)(2 y+x)}{x^{2}+y^{2}}=\frac{12 \times 576}{2880}=\frac{12}{5},
$$

i.e., $2 x^{2}-15 x y+22 y^{2}=0$,
Factoring, we get $(x-2 y)(2 x-11 y)=0$.
Since $2 x-11 y=0$ does not yield integer solutions for $x, y$, we discard it. From $x-2 y=0$, we get $x=2 y$, substituting into $x^{2}+y^{2}=2880$ yields $y=24$.

Therefore, $x=48, u=\frac{576}{2 \times 48-24}=8$. Thus, $A$ obtains 6 birds, and $B$ obtains 4 birds.