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

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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 integers 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>?
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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>
 
<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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== Solution ==
 
== Solution ==
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The sum is the highest if two factors are the lowest!
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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 \mathrm{(E)}</math>.
 
So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \mathrm{(E)}</math>.
  

Revision as of 11:10, 15 October 2006

In the year $2001$, the United States will host the International Mathematical Olympiad. Let $\displaystyle I,M,$ and $\displaystyle O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$. What is the largest possible value of the sum $\displaystyle I + M + O$?

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


Solution

The sum is the highest if two factors are the lowest! So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \mathrm{(E)}$.

See Also

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