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Difference between revisions of "2004 AIME I Problems/Problem 2"

 
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== Solution ==
 
== Solution ==
 +
Let us give the [[element]]s of our [[set]]s names:
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<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>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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<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>.
 +
 +
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 ==
 +
* [[2004 AIME I Problems/Problem 1| Previous problem]]
 +
 +
* [[2004 AIME I Problems/Problem 3| Next problem]]
 +
 
* [[2004 AIME I Problems]]
 
* [[2004 AIME I Problems]]

Revision as of 14:48, 18 August 2006

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 that $2 = x + \frac{m - 1}2$ and $x + (m - 1) = \frac{m + 3}2$ and also

$m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2$ and so $1 = 2y + (2m - 1)$ so $2m = 2(y + 2m - 1)$ and $m = y + 2m - 1$.

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

See also

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