Difference between revisions of "2000 AIME I Problems/Problem 10"
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== Problem == | == Problem == | ||
− | A sequence of numbers <math>x_{1},x_{2},x_{3},\ldots,x_{100}</math> has the property that, for every integer <math>k</math> between <math>1</math> and <math>100,</math> inclusive, the number <math>x_{k}</math> is <math>k</math> less than the sum of the other <math>99</math> numbers. Given that <math>x_{50} = m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m + n</math>. | + | A [[sequence]] of numbers <math>x_{1},x_{2},x_{3},\ldots,x_{100}</math> has the property that, for every [[integer]] <math>k</math> between <math>1</math> and <math>100,</math> inclusive, the number <math>x_{k}</math> is <math>k</math> less than the sum of the other <math>99</math> numbers. Given that <math>x_{50} = m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m + n</math>. |
== Solution == | == Solution == | ||
− | Let the sum of all of the terms in the sequence be <math>\mathbb{S}</math>. | + | Let the sum of all of the terms in the sequence be <math>\mathbb{S}</math>. Then for each integer <math>k</math>, <math>x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k</math>. Summing this up for all <math>k</math> from <math>1, 2, \ldots, 100</math>, |
− | < | + | <cmath>\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ |
+ | 100\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\ | ||
+ | \mathbb{S}&=\frac{2525}{49}\end{align*}</cmath> | ||
− | + | Now, substituting for <math>x_{50}</math>, we get <math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}</math>, and the answer is <math>75+98=\boxed{173}</math>. | |
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− | <math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49} | ||
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− | <math>75+98=\boxed{173}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=I|num-b=9|num-a=11}} | {{AIME box|year=2000|n=I|num-b=9|num-a=11}} | ||
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+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 11:46, 1 January 2008
Problem
A sequence of numbers has the property that, for every integer
between
and
inclusive, the number
is
less than the sum of the other
numbers. Given that
where
and
are relatively prime positive integers, find
.
Solution
Let the sum of all of the terms in the sequence be . Then for each integer
,
. Summing this up for all
from
,
Now, substituting for , we get
, and the answer is
.
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |