Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AIME I Problems/Problem 2"
 
(9 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
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>
+
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 ==
 +
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>.
+
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>.
−
 
 
  
 
== See also ==
 
== See also ==
−
* [[2006 AIME I Problems]]
+
{{AIME box|year=2006|n=I|num-b=1|num-a=3}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 +
{{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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_2&oldid=55203"