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Difference between revisions of "2004 AIME I Problems/Problem 2"
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== Problem ==
 
== Problem ==
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Set <math> A </math> consists of <math> m </math> consecutive integers whose sum is <math> 2m,  </math>and set <math> B </math> consists of <math> 2m </math> consecutive integers whose sum is <math> m. </math> The absolute value of the difference between the greatest element of <math> A </math> and the greatest element of <math> B </math> is 99. Find <math> m. </math>
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[[Set]] <math> A </math> consists of <math> m </math> consecutive integers whose sum is <math> 2m,  </math>and set <math> B </math> consists of <math> 2m </math> consecutive integers whose sum is <math> m. </math> The absolute value of the difference between the greatest element of <math> A </math> and the greatest element of <math> B </math> is <math>99</math>. Find <math> m. </math>
  
 
== Solution ==
 
== Solution ==
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Let us give the [[element]]s of our [[set]]s names:
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Let us give the [[element]]s of our sets names:
 
<math>A = \{x, x + 1, x + 2, \ldots, x + m - 1\}</math> and <math>B = \{y, y + 1, \ldots, y + 2m - 1\}</math>.  So we are given that
 
<math>A = \{x, x + 1, x + 2, \ldots, x + m - 1\}</math> and <math>B = \{y, y + 1, \ldots, y + 2m - 1\}</math>.  So we are given that
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<cmath>2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,</cmath>
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so <math>2 = x + \frac{m - 1}2</math> and <math>x + (m - 1) = \frac{m + 3}2</math>. Also,
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<cmath>m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,</cmath>
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so <math>1 = 2y + (2m - 1)</math> so <math>2m = 2(y + 2m - 1)</math> and <math>m = y + 2m - 1</math>.
  
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<math>2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2</math> so that <math>2 = x + \frac{m - 1}2</math> and <math>x + (m - 1) = \frac{m + 3}2</math> and also
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Then by the given, <math>99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|</math>.  <math>m</math> is a [[positive integer]] so we must have <math>99 = \frac{m - 3}2</math> and so <math>m = \boxed{201}</math>.
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<math>m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2</math> and so <math>1 = 2y + (2m - 1)</math> so <math>2m = 2(y + 2m - 1)</math> and <math>m = y + 2m - 1</math>.
 
−
 
 
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Then by the given, <math>99 = |(x + m - 1) - (y + 2m - 1)| = |\frac{m + 3}2 - m| = |\frac{m - 3}2|</math>.  <math>m</math> is a [[positive integer]] so we must have <math>99 = \frac{m - 3}2</math> and so <math>m = 201</math>.
 
  
 
== See also ==
 
== See also ==
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* [[2004 AIME I Problems/Problem 1| Previous problem]]
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{{AIME box|year=2004|n=I|num-b=1|num-a=3}}
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* [[2004 AIME I Problems/Problem 3| Next problem]]
 
  
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* [[2004 AIME I Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 12:26, 27 April 2008

Problem

Set $A$ consists of $m$ consecutive integers whose sum is $2m,$and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$. Find $m.$

Solution

Let us give the elements of our sets names: $A = \{x, x + 1, x + 2, \ldots, x + m - 1\}$ and $B = \{y, y + 1, \ldots, y + 2m - 1\}$. So we are given that \[2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,\] so $2 = x + \frac{m - 1}2$ and $x + (m - 1) = \frac{m + 3}2$. Also, \[m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,\] so $1 = 2y + (2m - 1)$ so $2m = 2(y + 2m - 1)$ and $m = y + 2m - 1$.

Then by the given, $99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|$. $m$ is a positive integer so we must have $99 = \frac{m - 3}2$ and so $m = \boxed{201}$.

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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