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 "2001 AIME II Problems/Problem 3"

(eqnarray -> align; LaTeX style)
(correct solution)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
Given that
 
Given that
 +
 
<cmath>
 
<cmath>
 
\begin{align*}x_{1}&=211,\\
 
\begin{align*}x_{1}&=211,\\
Line 8: Line 9:
 
x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
 
x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
 
</cmath>
 
</cmath>
 +
 
find the value of <math>x_{531}+x_{753}+x_{975}</math>.
 
find the value of <math>x_{531}+x_{753}+x_{975}</math>.
  
 
== Solution ==
 
== Solution ==
−
<math>x_5=x_4-x_3+x_2-x_1</math>
+
We find that <math>x_5 = 267</math> by the recursive formula. Summing the [[recursion]]s
−
 
 
−
<math>x_6=x_4-x_3+x_2-x_1-x_4+x_3-x_2=-x_1</math>
 
  
−
<math>x_7=-x_1-x_4+x_3-x_2+x_1+x_4-x_3=-x_2</math>
+
<cmath>\begin{align*}
 +
x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\
 +
x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5}
 +
\end{align*}</cmath>
  
−
<math>x_8=-x_2+x_1+x_4-x_3+x_2-x_1-x_4=-x_3</math>
+
yields <math>x_{n} = -x_{n-5}</math>. Thus <math>x_n = (-1)^k x_{n-5k}</math>. Since <math>531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5</math>, it follows that
  
−
<math>x_9=-x_3+x_2-x_1-x_4+x_3-x_2+x_1=-x_4</math>
+
<cmath>x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.</cmath>  
−
 
 
−
And it cycles back to <math>x_{11}=x_1</math>
 
−
 
 
−
Therefore, <math>x_{y}=x_{y-10}</math>, so
 
−
<cmath>
 
−
\begin{align*}x_{531}+x_{753}+x_{975}=x_1+x_3+x_5&=x_1+x_3+x_4-x_3+x_2-x_1\\
 
−
&=x_4+x_2=523+420=\boxed{943}\end{align*}
 
−
</cmath>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=II|num-b=2|num-a=4}}
 
{{AIME box|year=2001|n=II|num-b=2|num-a=4}}

Revision as of 21:18, 25 July 2008

Problem

Given that

\begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}

find the value of $x_{531}+x_{753}+x_{975}$.

Solution

We find that $x_5 = 267$ by the recursive formula. Summing the recursions

\begin{align*} x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\ x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5} \end{align*}

yields $x_{n} = -x_{n-5}$. Thus $x_n = (-1)^k x_{n-5k}$. Since $531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5$, it follows that

\[x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.\]

See also

2001 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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=2001_AIME_II_Problems/Problem_3&oldid=27193"