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"

m (rv)
Line 4: Line 4:
 
== 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 <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>.
 +
 +
Alternatively, for ease of calculation, let set <math>\mathcal{B}</math> be a 10-element subset of <math>\{1,2,3,\ldots,100\}, 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}.
  
 
== See also ==
 
== See also ==

Revision as of 06:39, 15 August 2011

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_2&oldid=41479"