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

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{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #1]] and [[2000 AMC 10 Problems|2000 AMC 10 #1]]}}
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==Problem==
 
==Problem==
In the year <math>2001</math>, the United States will host the [[International Mathematical Olympiad]]. Let <math> \displaystyle I,M,</math> and <math>\displaystyle 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>\displaystyle I + M + O</math>?
 
  
<math> \mathrm{(A) \ 23 } \qquad \mathrm{(B) \ 55 } \qquad \mathrm{(C) \ 99 } \qquad \mathrm{(D) \ 111 } \qquad \mathrm{(E) \ 671 </math>
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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>?
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<math>\text{(A)}\ 23 \qquad \text{(B)}\ 55 \qquad \text{(C)}\ 99 \qquad \text{(D)}\ 111 \qquad \text{(E)}\ 671</math>
  
 
== Solution ==
 
== Solution ==
The sum is the highest if two [[factor]]s are the lowest!
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So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \mathrm{(E)}</math>.
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The sum is the highest if two [[factor]]s are the lowest.
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So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \boxed{\text{(E)}}</math>.
  
 
==See Also==
 
==See Also==
{{AMC12 box|year=2000|before=First Question|num-a=2}}
 
  
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{{AMC12 box|year=2000|before=First<br />Question|num-a=2}}
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{{AMC10 box|year=2000|before=First<br />Question|num-a=2}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 15:36, 31 July 2009

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 $I + M + O$?

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

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 (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