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\}, 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}. | ||
== See also == | == See also == |
Revision as of 06:39, 15 August 2011
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
T
\mathcal{B}
S
T=5050-S
T
1+2+ \ldots +10 = 55
91+92+ \ldots + 100 = 955
955-55+1=\boxed{901}.
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 |