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Difference between revisions of "2006 AIME I Problems/Problem 2"
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== Solution ==
 
== Solution ==
  
−
The smallest S is <math>1+2+...+90=91\times45=4095</math>. The largest S is <math>11+12+...+100=111\times45=4995</math>. All numbers between 4095 and 4995 are possible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>.
+
The smallest S is <math>1+2+ \cdots +90=91\times45=4095</math>. The largest S is <math>11+12+ \cdots +100=111\times45=4995</math>. All numbers between 4095 and 4995 are possible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>.
  
  
 
== See also ==
 
== See also ==
 
* [[2006 AIME I Problems]]
 
* [[2006 AIME I Problems]]
 +
 +
[[Category:Intermediate Combinatorics Problems]]

Revision as of 17:10, 18 July 2006

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 $1+2+ \cdots +90=91\times45=4095$. The largest S is $11+12+ \cdots +100=111\times45=4995$. All numbers between 4095 and 4995 are possible values of S, so the number of possible values of S is $4995-4095+1=901$.


See also

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