Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 5: Line 5:
 
<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-2}=\dfrac{1}{m}-\dfrac{1}{n-2}</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 is
  
−
<math>\dfrac{n-2-m}{m(n-2)}=\dfrac{1}{29}</math>
+
<math>\dfrac{m-n}{mn}=\dfrac{1}{29}</math>
  
−
We cross-multiply to get
+
Since we need a 29 in the denominator, let <math>m=29t</math>*. Substituting,
  
−
<math>mn-2m=29n-58-29m\Rightarrow mn+27m-29n+58=0\Rightarrow (m-29)(n+27)=-725</math>
+
<math>29t-n=nt</math>
 +
<math>\frac{29t}{t+1} = n</math>
  
−
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>
+
Since n is an integer, <math>t+1 = 29</math>, or <math>t=28</math>. It quickly follows that <math>m=29(28)</math> and <math>n=28</math>, so <math>m+n = 30(28) = \fbox{840}</math>.
 +
 
 +
*If <math>n=29t</math>, a similar argument to the one above implies <math>n=29(28)</math> and <math>m=28</math>, which implies <math>m>n</math>, which is impossible since <math>m-n>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}}

Revision as of 16:13, 18 August 2008

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 is

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

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

$29t-n=nt$ $\frac{29t}{t+1} = n$

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

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

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_I_Problems/Problem_4&oldid=27472"