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 "2008 AMC 10B Problems/Problem 11"
m (→‎Solution)
 
(14 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileae for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four
+
Suppose that <math>(u_n)</math> is a [[sequence]] of real numbers satifying <math>u_{n+2}=2u_{n+1}+u_n</math>,
−
times. On December 31 the pedometer reads 50000. Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
+
 
 +
and that <math>u_3=9</math> and <math>u_6=128</math>. What is <math>u_5</math>?
 +
 
 +
<math>\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104</math>
  
−
(A) 2500 (B) 3000 (C) 3500 (D) 4000 (E) 4500
 
 
==Solution==
 
==Solution==
−
Every time the pedometer flips from <math>99999</math> to
+
If we plug in <math>n=4</math>, we get
−
 
 
−
<math>00000</math> Pete has walked <math>100000</math> steps.
 
  
−
So, if the pedometer flipped <math>44</math> times
+
<center><math>128=2u_5+u_4.</math></center>
  
−
Pete walked <math>44*100000+50000=4450000</math> steps.
+
By plugging in <math>n=3</math>, we get
  
−
Dividing by <math>1800</math> gives <math>2472.\overline{2}</math>
+
<center><math>u_5=2u_4+9.</math></center>
  
−
This is closest to answer <math>\boxed{A}</math>.
+
This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives <math>128=5u_4+18 \Longrightarrow u_4=22</math>, therefore <math>u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\boxed{(B)}}</math>.
 +
~ Mathkiddie
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2008|ab=B|num-b=10|num-a=12}}
 +
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:42, 24 October 2022

Problem

Suppose that $(u_n)$ is a sequence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$,

and that $u_3=9$ and $u_6=128$. What is $u_5$?

$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104$

Solution

If we plug in $n=4$, we get

$128=2u_5+u_4.$

By plugging in $n=3$, we get

$u_5=2u_4+9.$

This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives $128=5u_4+18 \Longrightarrow u_4=22$, therefore $u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\boxed{(B)}}$. ~ Mathkiddie

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 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=2008_AMC_10B_Problems/Problem_11&oldid=179450"