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: | ||
+ | <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 | ||
+ | |||
+ | <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 15:48, 18 August 2006
Problem
Set consists of
consecutive integers whose sum is
and set
consists of
consecutive integers whose sum is
The absolute value of the difference between the greatest element of
and the greatest element of
is 99. Find
Solution
Let us give the elements of our sets names:
and
. So we are given that
so that
and
and also
and so
so
and
.
Then by the given, .
is a positive integer so we must have
and so
.