2000 AMC 12 Problems/Problem 1: Difference between revisions

Latest revision as of 17:00, 6 November 2025

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 largest 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 $667 + 3 + 1 = \boxed{\text{(E) 671.}}$

~armang32324

Solution 2

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

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

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 $\boxed{\text{(E) 671}}$ as our final answer.

Solution 4 (Faster Way)

Notice $2001 = 3 * 29 * 23$ , so we can just maximize this with $667 * 3 * 1$ , which has a sum of $671$ . Our answer is $\boxed{\text{(E) 671}}$ .

-itsj

Video Solution by Power Solve

https://www.youtube.com/watch?v=YrAs8vadKMk

Video Solution by Daily Dose of Math

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

~Thesmartgreekmathdude

2000 AMC 12 (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

@@ Line 3: / Line 3: @@
 ==Problem==
-In the year <math>2001</math>, the United States will host the [[International Mathematical Olympiad]].  Let <math>I,M,</math> and <math>O</math> be distinct [[positive integer]]s such that the product <math>I \cdot M \cdot O = 2001 </math>.  What is the largest possible value of the sum <math>I + M + O</math>?
+In the year <imath>2001</imath>, the United States will host the [[International Mathematical Olympiad]].  Let <imath>I,M,</imath> and <imath>O</imath> be distinct [[positive integer]]s such that the product <imath>I \cdot M \cdot O = 2001 </imath>.  What is the largest possible value of the sum <imath>I + M + O</imath>?
-<math>\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671</math>
+<imath>\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671</imath>
 == Solution 1 (Verifying the Statement)==
-First, we need to recognize that a number is going to be largest only if, of the <math>3</math> [[factor]]s, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like <math>30</math>. It becomes much more clear that this is true, and in this situation, the value of <math>I + M + O</math> would be <math>18</math>. Now, we use this process on <math>2001</math> to get <math>667 * 3 * 1</math> as our <math>3</math> factors.
+First, we need to recognize that a number is going to be largest only if, of the <imath>3</imath> [[factor]]s, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like <imath>30</imath>. It becomes much more clear that this is true, and in this situation, the value of <imath>I + M + O</imath> would be <imath>18</imath>. Now, we use this process on <imath>2001</imath> to get <imath>667 * 3 * 1</imath> as our <imath>3</imath> factors.
-Hence, we have <math>667 + 3 + 1 = \boxed{\text{(E) 671.}}</math>
+Hence, we have <imath>667 + 3 + 1 = \boxed{\text{(E) 671.}}</imath>
 ~armang32324
@@ Line 17: / Line 17: @@
 The sum is the highest if two [[factor]]s are the lowest.
-So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \boxed{\text{(E) 671}}</math>.
+So, <imath>1 \cdot 3 \cdot 667 = 2001</imath> and <imath>1+3+667=671 \Longrightarrow \boxed{\text{(E) 671}}</imath>.
 == Solution 3 (Answer Choices) ==
-We see since <math>2 + 0 + 0 + 1</math> is divisible by <math>3</math>, we can eliminate all of the first <math>4</math> answer choices because they are way too small and get <math>\boxed{\text{(E) 671}}</math> as our final answer.
+We see since <imath>2 + 0 + 0 + 1</imath> is divisible by <imath>3</imath>, we can eliminate all of the first <imath>4</imath> answer choices because they are way too small and get <imath>\boxed{\text{(E) 671}}</imath> as our final answer.
 == Solution 4 (Faster Way) ==
-Notice <math>2001 = 3 * 29 * 23</math>, so we can just maximize this with <math>667 * 3 * 1</math>, which has a sum of <math>671</math>. Our answer is <math>\boxed{\text{(E) 671}}</math>.
+Notice <imath>2001 = 3 * 29 * 23</imath>, so we can just maximize this with <imath>667 * 3 * 1</imath>, which has a sum of <imath>671</imath>. Our answer is <imath>\boxed{\text{(E) 671}}</imath>.
 -itsj

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