[Number Theory]
Problem: Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. For example, $\lfloor 3.1\rfloor=3$ and $\lfloor-1.4\rfloor=-2$.

Suppose that $f(n)=2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ and $g(n)=2 n+\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ for each positive integer $n$.
Determine a value of $n$ for which $f(n)=100$.

Solution: To determine a value of $n$ for which $f(n)=100$, we need to solve the equation

$$
2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor=100
$$

We first solve the equation

$$
2 x-\frac{1+\sqrt{8 x-7}}{2}=100 \quad(* *)
$$

because the left sides of $(*)$ and $(* *)$ do not differ by much and so the solutions are likely close together. We will try integers $n$ in $(*)$ that are close to the solutions to $(* *)$.

Manipulating $(* *)$, we obtain

$$
\begin{aligned}
4 x-(1+\sqrt{8 x-7}) & =200 \\
4 x-201 & =\sqrt{8 x-7} \\
(4 x-201)^{2} & =8 x-7 \\
16 x^{2}-1608 x+40401 & =8 x-7 \\
16 x^{2}-1616 x+40408 & =0 \\
2 x^{2}-202 x+5051 & =0
\end{aligned}
$$

By the quadratic formula,

$$
x=\frac{202 \pm \sqrt{202^{2}-4(2)(5051)}}{2(2)}=\frac{202 \pm \sqrt{396}}{4}=\frac{101 \pm \sqrt{99}}{2}
$$

and so $x \approx 55.47$ or $x \approx 45.53$.

We try $n=55$, which is close to 55.47 :

$$
f(55)=2(55)-\left\lfloor\frac{1+\sqrt{8(55)-7}}{2}\right\rfloor=110-\left\lfloor\frac{1+\sqrt{433}}{2}\right\rfloor
$$

Since $\sqrt{433} \approx 20.8$, then $\frac{1+\sqrt{433}}{2} \approx 10.9$, which gives $\left\lfloor\frac{1+\sqrt{433}}{2}\right\rfloor=10$.

Thus, $f(55)=110-10=100$.

Therefore, a value of $n$ for which $f(n)=100$ is $n=55$.

Final answer: 55