2000 AMC 12 Problems/Problem 1

The following problem is from both the 2000 AMC 12 #1 and 2000 AMC 10 #1, so both problems redirect to this page.

Problem

In the year $2001$, the United States will host the International Mathematical Olympiad. Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$. What is the largest possible value of the sum $I + M + O$?

$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$

Solution 1 (Verifying the Statement)

First, we need to recognize that a number is going to be lowest only if, of the $3$ factors, 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

Solution 2

The sum is the highest if two factors are the lowest.

So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \boxed{\text{(E)}}$.

Solution 3 (Answer Choices)

We see since $2 + 0 + 0 + 1$ is divisible by $3$, we can eliminate all of the first $4$ answer choices because they are way too small and get $\boxed{\text{E}}$ as our final answer.

See Also

2000 AMC 12 (ProblemsAnswer KeyResources)
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First
Question
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Problem 2
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All AMC 12 Problems and Solutions
2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
First
Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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