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 17:34, 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 |