2000 AMC 10 Problem 1

1. First break $2001$ into primes. Its digits add to $2 + 0 + 0 + 1 = 3$, so $3$ divides it: $2001 = 3 \cdot 667$. Trying small primes on $667$ gives $667 = 23 \cdot 29$. So
$$2001 = 3 \cdot 23 \cdot 29.$$
2. Now think about what makes a sum big when the product is fixed. Compare $36 = 6 \cdot 6$ (sum $12$) with $36 = 1 \cdot 36$ (sum $37$). Lopsided numbers give a much bigger sum. So we want two tiny numbers and one huge one.
3. The three numbers must be different positive whole numbers. The smallest one we can use is $1$. The next smallest number that divides $2001$ is $3$. That leaves $2001 \div 3 = 667$ for the third number.
4. The sum is $1 + 3 + 667 = 671$.
5. To be safe, list every other way to write $2001$ as a product of three different numbers:

The first row is the largest.

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

When the product of some numbers is fixed, their sum gets bigger as the numbers get more spread out. So the best choice uses the smallest possible numbers, and $1$ counts. The word "different" only stops you from using $1, 1, 2001$.

Use the answer choices. Once you see $2001 = 3 \cdot 667$, the numbers $1, 3, 667$ are allowed and add to $671$. That is the biggest choice offered, and it can actually be reached, so it must be the answer.

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 $\boxed{\textbf{(E)}\ 671}$.