Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See Also)
(→‎Problem)
Line 5: Line 5:
 
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 <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>?
  
−
<math>\text{(A)}\ 23 \qquad \text{(B)}\ 55 \qquad \text{(C)}\ 99 \qquad \text{(D)}\ 111 \qquad \text{(E)}\ 671</math>
+
<math>\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 20:18, 23 November 2020

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

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

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

See Also

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_1&oldid=138312"