1. We are given the function $$f(n) = \frac{(\frac{100}{3})^n}{n!}$$ and need to find the positive integer $n$ for which $f(n)$ is largest. 2. The factorial $n! = n \times (n-1) \times \cdots \times 1$ grows very quickly, so there's a point where increasing $n$ no longer increases $f(n)$. 3. To find where $f(n)$ is largest, consider the ratio: $$\frac{f(n+1)}{f(n)} = \frac{(\frac{100}{3})^{n+1}/(n+1)!}{(\frac{100}{3})^n/n!} = \frac{\frac{100}{3}}{n+1}$$ 4. We look for the integer $n$ where $f(n+1) \leq f(n)$, i.e. where the ratio falls below or equals 1: $$\frac{100/3}{n+1} \leq 1 \implies n+1 \geq \frac{100}{3} \implies n+1 \geq 33.33...$$ So the maximal $f(n)$ occurs when $n$ is just before the ratio falls below 1; that is when $n+1=34$ or $n=33$. Then: - For $n=33$, $\frac{f(34)}{f(33)} = \frac{100/3}{34} \approx 0.98 < 1$ (function starts decreasing here), - For $n=32$, $\frac{f(33)}{f(32)} = \frac{100/3}{33} \approx 1.01 > 1$ (function still increasing). 5. Therefore, the maximum value of $f(n)$ is at $n=33$. Final answer: (E) 33