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

Revision as of 12:56, 25 September 2007

Problem

Let set $\mathcal{A}$ be a 90-element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$

Solution

By the Triangle Inequality:

$\log_{10} 12 + \log_{10} n > \log_{10} 75$

$\log_{10} 12n > \log_{10} 75$

$12n > 75$

$n > \frac{75}{12} = \frac{25}{4} = 6.25$

Also:

$\log_{10} 12 + \log_{10} 75 > \log_{10} n$

$\log_{10} 12\cdot75 > \log_{10} n$

$n < 900$

Combining these two inequalities:

$6.25 < n < 900$

The number of possible integer values for $n$ is the number of integers over the interval $(6.25 , 900)$ , which is $893$ .

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 == Problem ==
-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>.
+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>
 == Solution ==

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

Revision as of 12:56, 25 September 2007

Problem

Solution

See also