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

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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 <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}.
+
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 ==

Revision as of 05:43, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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