2006 AIME I Problems/Problem 2: Difference between revisions
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== 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+ \ | |||
== See also == | == See also == | ||
{{AIME box|year=2006|n=I|num-b=1|num-a=3}} | |||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 20:21, 11 March 2007
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 is 
. The largest is 
. All numbers between 
and 
are possible values of S, so the number of possible values of S is 
.
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 | ||
