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"

 
(→‎Non-Rigorous Solution)
 
(9 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Given that
 +
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
find the value of <math>x_{531}+x_{753}+x_{975}</math>.
  
 
== Solution ==
 
== Solution ==
 +
We find that <math>x_5 = 267</math> by the recursive formula. Summing the [[recursion]]s
 +
 +
<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>
 +
 +
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
 +
 +
<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>
 +
 +
== Solution Variant ==
 +
The recursive formula suggests telescoping. Indeed, if we add <math>x_n</math> and <math>x_{n-1}</math>, we have <math>x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}</math>.
 +
 +
Subtracting <math>x_{n-1}</math> yields <math>x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}</math>.
 +
 +
Thus,
 +
 +
<cmath>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_2 + x_4 = 375 + 523 = \boxed{898}.</cmath>
 +
 +
Notice that we didn't need to use the values of <math>x_1</math> or <math>x_3</math> at all.
 +
 +
 +
==Non-Rigorous Solution==
 +
 +
Calculate the first few terms:
 +
 +
<cmath>211,375,420,523,267,-211,-375,-420,-523,\dots</cmath>
 +
 +
At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously <math>211+420+267=\boxed{898}</math>
 +
 +
~Dhillonr25
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/lH-0ul1hwKw?t=870
 +
 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==
−
* [[2001 AIME II Problems]]
+
{{AIME box|year=2001|n=II|num-b=2|num-a=4}}
 +
{{MAA Notice}}

Latest revision as of 00:28, 12 November 2022

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

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

Solution Variant

The recursive formula suggests telescoping. Indeed, if we add $x_n$ and $x_{n-1}$, we have $x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}$.

Subtracting $x_{n-1}$ yields $x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}$.

Thus,

\[x_{531} + x_{753} + x_{975} = x_1 + x_3 + x_5 = x_1 + x_3 + (x_4 - x_3 + x_2 - x_1) = x_2 + x_4 = 375 + 523 = \boxed{898}.\]

Notice that we didn't need to use the values of $x_1$ or $x_3$ at all.


Non-Rigorous Solution

Calculate the first few terms:

\[211,375,420,523,267,-211,-375,-420,-523,\dots\]

At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously $211+420+267=\boxed{898}$

~Dhillonr25

Video Solution by OmegaLearn

https://youtu.be/lH-0ul1hwKw?t=870

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_3&oldid=180825"