Difference between revisions of "2006 AIME I Problems/Problem 2"
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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 <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>. | ||
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| + | 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 == | ||
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:05, 4 July 2013
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 
.
Alternatively, for ease of calculation, let set 
be a 10-element subset of 
, and let 
be the sum of the elements of . Note that the number of possible is the number of possible 
. The smallest possible is 
and the largest is 
, so the number of possible values of T, and therefore 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
