1. Find the sum of all positive integers whose largest proper divisor is 55 . (A proper divisor of $n$ is divisor that is strictly less than $n$.)
Proposed by: Michael Tang
Answer: $\square$
The largest proper divisor of an integer $n$ is $\frac{n}{p}$, where $p$ is the smallest prime divisor of $n$. So $n=55 p$ for some prime $p$. Since $55=5 \cdot 11$, we must have $p \leq 5$, so $p=2,3,5$ gives all solutions. The sum of these solutions is $55(2+3+5)=550$.