1. Given an integer $n \geq 2$, let $f(n)$ be the second largest positive divisor of $n$. For example, $f(12)=6$ and $f(13)=1$. Determine the largest positive integer $n$ such that $f(n)=35$.
Solution: For a positive integer $n$, let $n^{\prime}$ be the smallest prime that divides $n$. Then we can see that $f(n)=n / n^{\prime}$ and also that $n=f(n) \cdot n^{\prime}$. We are given that $f(n)=35$, so to maximize $n$ we must maximize $n^{\prime}$. Since $n^{\prime}$ is the smallest prime factor of $n$, it cannot be larger than any factor of $f(n)$. Thus, the largest possible value of $n^{\prime}$ is 5 , so the largest possible value of $n$ is $5 \cdot 35=175$.