Difference between revisions of "2008 AMC 10B Problems/Problem 11"
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(A) 2500 (B) 3000 (C) 3500 (D) 4000 (E) 4500 | (A) 2500 (B) 3000 (C) 3500 (D) 4000 (E) 4500 | ||
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| + | ==Problem== | ||
| + | {{Suppose that <math>(u_n)</math> is a serquence of real numbers satifying <math>u_{n+2}=2u_{n+1}+u_n</math>, | ||
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| + | and that <math>u_3=9</math> and <math>u_6=128</math>. What is <math>u_5</math>? | ||
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| + | (A) 40 (B) 53 (C) 68 (D) 88 (E) 104}} | ||
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==Solution== | ==Solution== | ||
Every time the pedometer flips from <math>99999</math> to | Every time the pedometer flips from <math>99999</math> to | ||
Revision as of 16:34, 10 August 2008
Contents
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 
is a serquence of real numbers satifying 
,
and that 
and 
. What is 
?
(A) 40 (B) 53 (C) 68 (D) 88 (E) 104}}
Solution
Every time the pedometer flips from 
to
Pete has walked 
steps.
So, if the pedometer flipped 
times
Pete walked 
steps.
Dividing by 
gives 
This is closest to answer 
.
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 | ||
