In the year $2001$, the United States will host the [International Mathematical Olympiad](https://artofproblemsolving.com/wiki/index.php/International_Mathematical_Olympiad).  Let $I,M,$ and $O$ be distinct [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) such that the product $I \cdot M \cdot O = 2001$.  What is the largest possible value of the sum $I + M + O$?
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$
First, we need to recognize that a number is going to be lowest only if, of the $3$ [factors](https://artofproblemsolving.com/wiki/index.php/Factor), two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$. It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be $18$. Now, we use this process on $2001$ to get $667 * 3 * 1$ as our $3$ factors.
Hence, we have $667 + 3 + 1 = \boxed{\text{(E) 671}}$
Solution By: armang32324