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

(Problem)
(Solution)
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==Solution==
 
==Solution==
Every time the pedometer flips from <math>99999</math> to
+
{{We konw that <math>u_6=128</math>, so we plug in <math>n=4</math> to get <math>128=2u_5+u_4</math>. We plug in <math>n=3</math> to get
  
<math>00000</math> Pete has walked <math>100000</math> steps.
+
<math>u_5=2u_4+9</math>. Substituting gives
  
So, if the pedometer flipped <math>44</math> times
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<math>128=5u_4+18 \rightarrow u_4=22</math>
  
Pete walked <math>44*100000+50000=4450000</math> steps.
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This gives <math>u_5=\frac{128-22}{2}=53</math>.
  
Dividing by <math>1800</math> gives <math>2472.\overline{2}</math>
+
Answer B is the correct answer
  
This is closest to answer <math>\boxed{A}</math>.
+
NOTE: This is my (BOGTRO) solution, not the official one,
 +
 
 +
and should be ignored in view of a better solution.}}
 +
 
 +
==See also==
 +
{{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}}
  
 
==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}}

Revision as of 17:34, 10 August 2008

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

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

(A) 40 (B) 53 (C) 68 (D) 88 (E) 104}}

Solution

{{We konw that $u_6=128$, so we plug in $n=4$ to get $128=2u_5+u_4$. We plug in $n=3$ to get

$u_5=2u_4+9$. Substituting gives

$128=5u_4+18 \rightarrow u_4=22$

This gives $u_5=\frac{128-22}{2}=53$.

Answer B is the correct answer

NOTE: This is my (BOGTRO) solution, not the official one,

and should be ignored in view of a better solution.}}

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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

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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