Difference between revisions of "2000 AMC 10 Problems/Problem 1"

(New page: 2001=1*3*667=3*23*29 1+3+667=671 3+23+29=55 1+29+69=99 1+23+87=111 Clearly, 671, or E is the largest.)
 
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2001=1*3*667=3*23*29
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==Problem==
  
1+3+667=671
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In the year 2001, the United States will host the International Mathematical Olympiad. Let <math>I</math>, <math>M</math>, and <math>O</math> be distinct positive integers 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>?
3+23+29=55
 
1+29+69=99
 
1+23+87=111
 
  
Clearly, 671, or E is the largest.
+
==Solution==
 +
 
 +
<math>2001=1\cdot 3\cdot 667=3\cdot 23\cdot 29</math>
 +
 
 +
<math>1+3+667=671</math>
 +
 
 +
<math>3+23+29=55</math>
 +
 
 +
<math>1+29+69=99</math>
 +
 
 +
<math>1+23+87=111</math>
 +
 
 +
Clearly, <math>671</math>, or <math>\boxed{E}</math> is the largest.

Revision as of 14:07, 7 January 2009

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 $I+M+O$?

Solution

$2001=1\cdot 3\cdot 667=3\cdot 23\cdot 29$

$1+3+667=671$

$3+23+29=55$

$1+29+69=99$

$1+23+87=111$

Clearly, $671$, or $\boxed{E}$ is the largest.