Difference between revisions of "2002 AIME I Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | {{ | + | <math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus, |
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| + | <math>a_m+a_{m+1}+a_{m+2}+\cdots +a_{n-1}=\dfrac{1}{m}-\dfrac{1}{m+1}+\dfrac{1}{m+1}-\dfrac{1}{m+2}+\cdots +\dfrac{1}{n-1}-\dfrac{1}{n-2}=\dfrac{1}{m}-\dfrac{1}{n-2}</math> | ||
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| + | Which is | ||
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| + | <math>\dfrac{n-2-m}{m(n-2)}=\dfrac{1}{29}</math> | ||
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| + | We cross-multiply to get | ||
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| + | <math>mn-2m=29n-58-29m\Rightarrow mn+27m-29n+58=0\Rightarrow (m-29)(n+27)=-725</math> | ||
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| + | Thus <math>m</math> is an integer less than 29. We try <math>m=4</math> to get <math>n=2</math>. <math>m=24, n=118</math>. <math>m+n=\boxed{142}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=3|num-a=5}} | {{AIME box|year=2002|n=I|num-b=3|num-a=5}} | ||
Revision as of 08:07, 5 May 2008
Problem
Consider the sequence defined by 
for 
. Given that 
, for positive integers 
and 
with 
, find 
.
Solution
. Thus,
Which is
We cross-multiply to get
Thus is an integer less than 29. We try 
to get 
. 
. 
See also
| 2002 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
