Art of Problem Solving

2006 AIME I Problems/Problem 2: Difference between revisions

Joml88 (talk | contribs)
No edit summary
Joml88 (talk | contribs)
No edit summary
Line 9: Line 9:
== See also ==
== See also ==
* [[2006 AIME I Problems]]
* [[2006 AIME I Problems]]
* [[Combinatorics]]


[[Category:Intermediate Combinatorics Problems]]
[[Category:Intermediate Combinatorics Problems]]

Revision as of 16:15, 18 July 2006

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 $1+2+ \cdots +90=91\times45=4095$. The largest S is $11+12+ \cdots +100=111\times45=4995$. 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