Difference between revisions of "2006 AIME I Problems/Problem 2"

Revision as of 12:56, 25 September 2007

Problem

The lengths of the sides of a triangle with positive area are $\log_{10} 12$ , $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$ .

Solution

The smallest $S$ is $1+2+ \ldots +90 = 91 \cdot 45 = 4095$ . The largest $S$ is $11+12+ \ldots +100=111\cdot 45=4995$ . All numbers between $4095$ and $4995$ are possible values of S, so the number of possible values of S is $4995-4095+1=901$ .

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 == Problem ==
-Let [[set]] <math> \mathcal{A} </math> be a 90-[[element]] [[subset]] of <math> \{1,2,3,\ldots,100\}, </math> and let <math> S </math> be the sum of the elements of <math> \mathcal{A}. </math> Find the number of possible values of <math> S. </math>
+The lengths of the sides of a triangle with positive area are <math>\log_{10} 12</math>, <math>\log_{10} 75</math>, and <math>\log_{10} n</math>, where <math>n</math> is a positive integer. Find the number of possible values for <math>n</math>.
 == Solution ==

2006 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2006 AIME I Problems/Problem 2"

Revision as of 12:56, 25 September 2007

Problem

Solution

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