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Difference between revisions of "2006 AIME I Problems/Problem 2"
 
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== Problem ==
 
== Problem ==
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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>
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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 ==
 
== 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>.
  
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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>.
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Alternatively, for ease of calculation, let set <math>\mathcal{B}</math> be a 10-element subset of <math>\{1,2,3,\ldots,100\}</math>, and let <math>T</math> be the sum of the elements of <math>\mathcal{B}</math>. Note that the number of possible <math>S</math> is the number of possible <math>T=5050-S</math>. The smallest possible <math>T</math> is <math>1+2+ \ldots +10 = 55</math> and the largest is <math>91+92+ \ldots + 100 = 955</math>, so the number of possible values of T, and therefore S, is <math>955-55+1=\boxed{901}</math>.
  
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== See also ==
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{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
  
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== See also ==
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[[Category:Intermediate Combinatorics Problems]]
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* [[2006 AIME I Problems]]
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{{MAA Notice}}

Latest revision as of 19:05, 4 July 2013

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 $4995-4095+1=901$.

Alternatively, for ease of calculation, let set $\mathcal{B}$ be a 10-element subset of $\{1,2,3,\ldots,100\}$, and let $T$ be the sum of the elements of $\mathcal{B}$. Note that the number of possible $S$ is the number of possible $T=5050-S$. The smallest possible $T$ is $1+2+ \ldots +10 = 55$ and the largest is $91+92+ \ldots + 100 = 955$, so the number of possible values of T, and therefore S, is $955-55+1=\boxed{901}$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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