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Difference between revisions of "2002 AIME I Problems/Problem 4"

(→‎Solution)
m (Made the asterisk note at the end more obvious that it refers to the asterisk, got rid of that weird extra plus, and changed the wording a tiny bit)
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<math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus,
 
<math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus,
  
−
<math>a_m+a_{m+1}++\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}=\dfrac{1}{m}-\dfrac{1}{n}</math>
+
<math>a_m+a_{m+1}+\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}=\dfrac{1}{m}-\dfrac{1}{n}</math>
  
−
Which is
+
Which means
  
 
<math>\dfrac{n-m}{mn}=\dfrac{1}{29}</math>
 
<math>\dfrac{n-m}{mn}=\dfrac{1}{29}</math>
  
−
Since we need a 29 in the denominator, let <math>n=29t</math>.* Substituting,
+
Since we need a 29 in the denominator, we let <math>n=29t</math>.* Substituting, we get
  
 
<math>29t-m=mt</math>
 
<math>29t-m=mt</math>
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Since m is an integer, <math>t+1 = 29</math>, or <math>t=28</math>. It quickly follows that <math>n=29(28)</math> and <math>m=28</math>, so <math>m+n = 30(28) = \fbox{840}</math>.
 
Since m is an integer, <math>t+1 = 29</math>, or <math>t=28</math>. It quickly follows that <math>n=29(28)</math> and <math>m=28</math>, so <math>m+n = 30(28) = \fbox{840}</math>.
  
−
*If <math>m=29t</math>, a similar argument to the one above implies <math>m=29(28)</math> and <math>n=28</math>, which implies <math>m>n</math>. This is impossible since <math>n-m>0</math>.
+
*If <math>m=29t</math>, a similar argument to the one above implies <math>m=29(28)</math> and <math>n=28</math>, which implies <math>m>n</math>. This is impossible since <math>n-m>0</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:13, 11 July 2017

Problem

Consider the sequence defined by $a_k =\dfrac{1}{k^2+k}$ for $k\geq 1$. Given that $a_m+a_{m+1}+\cdots+a_{n-1}=\dfrac{1}{29}$, for positive integers $m$ and $n$ with $m<n$, find $m+n$.

Solution

$\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$. Thus,

$a_m+a_{m+1}+\cdots +a_{n-1}=\dfrac{1}{m}-\dfrac{1}{m+1}+\dfrac{1}{m+1}-\dfrac{1}{m+2}+\cdots +\dfrac{1}{n-1}-\dfrac{1}{n}=\dfrac{1}{m}-\dfrac{1}{n}$

Which means

$\dfrac{n-m}{mn}=\dfrac{1}{29}$

Since we need a 29 in the denominator, we let $n=29t$.* Substituting, we get

$29t-m=mt$

$\frac{29t}{t+1} = m$

Since m is an integer, $t+1 = 29$, or $t=28$. It quickly follows that $n=29(28)$ and $m=28$, so $m+n = 30(28) = \fbox{840}$. 

*If $m=29t$, a similar argument to the one above implies $m=29(28)$ and $n=28$, which implies $m>n$. This is impossible since $n-m>0$.

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

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