Difference between revisions of "2000 AMC 12 Problems/Problem 1"

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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
 
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== Video Solution (Daily Dose of Math) ==
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https://www.youtube.com/watch?v=aSzsStkkYeA
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--THESMARTGREEKMATHDUDE

Revision as of 15:12, 14 July 2024

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)
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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Video Solution (Daily Dose of Math)

https://www.youtube.com/watch?v=aSzsStkkYeA

--THESMARTGREEKMATHDUDE