2006 AIME I Problems/Problem 2: Difference between revisions
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== Solution == | == Solution == | ||
The smallest S is <math>1+2+ | 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>. | ||
== See also == | == See also == | ||
* [[2006 AIME I Problems]] | * [[2006 AIME I Problems]] | ||
[[Category:Intermediate Combinatorics Problems]] | |||
Revision as of 16:10, 18 July 2006
Problem
Let set 
be a 90-element subset of 
and let 
be the sum of the elements of 
Find the number of possible values of 
Solution
The smallest S is 
. The largest S is 
. All numbers between 4095 and 4995 are possible values of S, so the number of possible values of S is 
.
