[Number Theory]
Problem: For every real number $x$, define $\lfloor x\rfloor$ to be equal to the greatest integer less than or equal to $x$. (We call this the "floor" of $x$.) For example, $\lfloor 4.2\rfloor=4,\lfloor 5.7\rfloor=5$, $\lfloor-3.4\rfloor=-4,\lfloor 0.4\rfloor=0$, and $\lfloor 2\rfloor=2$.
Determine the integer equal to $\left\lfloor\frac{1}{3}\right\rfloor+\left\lfloor\frac{2}{3}\right\rfloor+\left\lfloor\frac{3}{3}\right\rfloor+\ldots+\left\lfloor\frac{59}{3}\right\rfloor+\left\lfloor\frac{60}{3}\right\rfloor$. (The sum has 60 terms.)

Solution: Since $0<\frac{1}{3}<\frac{2}{3}<1$, then $\left\lfloor\frac{1}{3}\right\rfloor=\left\lfloor\frac{2}{3}\right\rfloor=0$.

Since $1 \leq \frac{3}{3}<\frac{4}{3}<\frac{5}{3}<2$, then $\left\lfloor\frac{3}{3}\right\rfloor=\left\lfloor\frac{4}{3}\right\rfloor=\left\lfloor\frac{5}{3}\right\rfloor=1$.

These fractions can continue to be grouped in groups of 3 with the last full group of 3 satisfying $19 \leq \frac{57}{3}<\frac{58}{3}<\frac{59}{3}<20$, which means that $\left\lfloor\frac{57}{3}\right\rfloor=\left\lfloor\frac{58}{3}\right\rfloor=\left\lfloor\frac{59}{3}\right\rfloor=19$.

The last term is $\left\lfloor\frac{60}{3}\right\rfloor=\lfloor 20\rfloor=20$.



If the given sum is $S$, we obtain

$$
\begin{aligned}
S & =2 \cdot 0+3 \cdot 1+3 \cdot 2+\cdots+3 \cdot 19+1 \cdot 20 \\
& =0+3(1+2+\cdot+19)+20 \\
& =3 \cdot \frac{1}{2} \cdot 19 \cdot 20+20 \\
& =570+20 \\
& =590
\end{aligned}
$$

Final answer: 590