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

Revision as of 13:56, 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

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 $4995-4095+1=901$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-By the [[Triangle Inequality]]:
+The smallest <math>S</math> is <math>1+2+ \ldots +90 = 91 \cdot 45 = 4095</math>. The largest <math>S</math> is <math>11+12+ \ldots +100=111\cdot 45=4995</math>. All numbers between <math>4095</math> and <math>4995</math> are possible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>.
-<math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
-<math>\log_{10} 12n > \log_{10} 75 </math>
-<math> 12n > 75 </math>
-<math> n > \frac{75}{12} = \frac{25}{4} = 6.25 </math>
-Also:
-<math>\log_{10} 12 + \log_{10} 75 > \log_{10} n </math>
-<math>\log_{10} 12\cdot75 > \log_{10} n </math>
-<math> n < 900 </math>
-Combining these two inequalities:
-<math> 6.25 < n < 900 </math>
-The number of possible integer values for <math>n</math> is the number of integers over the interval <math>(6.25 , 900)</math>, which is <math>893</math>.
 == See also ==

Page

Toolbox

Search

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

Revision as of 13:56, 25 September 2007

Problem

Solution

See also