Difference between revisions of "2008 AMC 10B Problems/Problem 11"

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==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
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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?
 
 
 
(A) 2500 (B) 3000 (C) 3500 (D) 4000 (E) 4500
 
 
 
==Problem==
 
{{Suppose that <math>(u_n)</math> is a serquence of real numbers satifying <math>u_{n+2}=2u_{n+1}+u_n</math>,
 
  
 
and that <math>u_3=9</math> and <math>u_6=128</math>. What is <math>u_5</math>?
 
and that <math>u_3=9</math> and <math>u_6=128</math>. What is <math>u_5</math>?
  
(A) 40 (B) 53 (C) 68 (D) 88 (E) 104}}
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<math>\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104</math>
  
 
==Solution==
 
==Solution==
Every time the pedometer flips from <math>99999</math> to
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Plugging 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
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<center><math>128=2u_5+u_4.</math></center>
  
Pete walked <math>44*100000+50000=4450000</math> steps.
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Plugging in <math>n=3</math>, we get
  
Dividing by <math>1800</math> gives <math>2472.\overline{2}</math>
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<center><math>u_5=2u_4+9.</math></center>
  
This is closest to answer <math>\boxed{A}</math>.
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This is simply a system of two equations with two unknowns. Substituting gives <math>128=5u_4+18 \Longrightarrow u_4=22</math>, and <math>u_5=\frac{128-22}{2}=53 \longleftarrow \textbf{(B)}</math>.
  
 
==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}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 21:45, 13 November 2017

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

Plugging in $n=4$, we get

$128=2u_5+u_4.$

Plugging in $n=3$, we get

$u_5=2u_4+9.$

This is simply a system of two equations with two unknowns. Substituting gives $128=5u_4+18 \Longrightarrow u_4=22$, and $u_5=\frac{128-22}{2}=53 \longleftarrow \textbf{(B)}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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

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