16. Let $\lfloor x\rfloor$ be the greatest integer smaller than or equal to $x$. How many solutions are there to the equation $x^{3}-\left\lfloor x^{3}\right\rfloor=(x-\lfloor x\rfloor)^{3}$ on the interval $[1,20]$ ?
16. Answer: 9241.

Let $n=\lfloor x\rfloor,\{x\}=x-n$. The equation becomes $(n+\{x\})^{3}-\{x\}^{3}=\left\lfloor(n+\{x\})^{3}\right\rfloor$. Then
$$
3 n\{x\}(n+\{x\})=\left\lfloor 3 n\{x\}(n+\{x\})+\{x\}^{3}\right\rfloor .
$$

The right-hand side is an integer. The above holds if and only if $3 n\{x\}(n+\{x\})$ is an integer.
Note that $0 \leq 3 n\{x\}(n+\{x\})<3 n(n+1)$. There are exactly $3 n(n+1)$ solutions in $[n, n+1)$, $n=1,2, \ldots$. So on $[1,20]$, the total number of solutions is
$$
\begin{aligned}
& 3(1 \times 2+2 \times 3+\cdots 20 \times 21)+1 \\
= & \left(2^{3}-1^{3}\right)+\left(3^{3}-2^{3}\right)+\cdots+\left(21^{3}-20^{3}\right)-20+1 \\
= & 21^{3}-20=9241 .
\end{aligned}
$$