2000 AMC 10 Problem 1

The digit sum of $2001$ is $3$, so $3$ divides it: $2001 = 3 \cdot 667 = 3 \cdot 23 \cdot 29$.

We want three distinct positive integers whose product is $2001$ and whose sum is as large as possible. For a fixed product, a sum grows as the factors become more lopsided, so we want one giant factor and two tiny ones. The two smallest distinct factors we can use are $1$ and $3$, which leaves $667$ for the third:
$$1 + 3 + 667 = 671.$$

For comparison, the other ways to split $2001$ into three distinct factors give $1 + 23 + 87 = 111$, $1 + 29 + 69 = 99$, and $3 + 23 + 29 = 55$, all smaller.

The answer is $\boxed{\textbf{(E)}\ 671}$.

When a product is fixed, spreading the factors apart increases their sum (compare $1 + 36$ with $6 + 6$). The problem's only real content is the factorization of $2001$ and remembering that $1$ counts as a positive integer factor; "distinct" merely rules out $1, 1, 2001$.

The choices are ordered, and $671 = 667 + 3 + 1$ is visibly the sum of three factors of $2001$ once you check $3 \cdot 667 = 2001$. Since it is the largest option offered and it is attainable, it must be the answer.

• Using the three prime factors 3, 23, 29 (sum 55) without noticing that 1 is an allowed factor.
• Failing to test $2001$ for divisibility by $3$ and treating it as prime, which leads nowhere because $1, 1, 2001$ is not distinct.