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

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== Solution ==
 
== Solution ==
 
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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>.
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/Problem 1 | Previous problem]]
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{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
* [[2006 AIME I Problems/Problem 3 | Next problem]]
 
* [[2006 AIME I Problems]]
 
* [[Combinatorics]]
 
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 21:21, 11 March 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 $4995-4095+1=901$.

See also

2006 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions