Difference between revisions of "2009 AIME I Problems/Problem 7"
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<cmath>a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}</cmath> | <cmath>a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}</cmath> | ||
<cmath>a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}</cmath> | <cmath>a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}</cmath> | ||
− | Since <math>a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}</math>, we can easily use induction to show that <math>a_n = \log_5{(3n + 2)}</math>. So now we only need to find the next value of <math>n</math> that makes <math>\log_5{(3n + 2)}</math> an integer. This means that <math>3n + 2</math> must be a power of <math>5</math>. We test <math>25</math>: <cmath>3n + 2 = 25</cmath> <cmath>3n = 23</cmath> This has no integral solutions, so we try <math>125</math>: <cmath>3n + 2 = 125</cmath> <cmath>3n = 123</cmath> <cmath>n = \boxed{041}</cmath> | + | Plug in <math>n = 1, 2, 3, 4</math> to see the first few terms of the sequence: <cmath>\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.</cmath> We notice that the terms <math>5, 8, 11, 14</math> are in arithmetic progression. Since <math>a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}</math>, we can easily use induction to show that <math>a_n = \log_5{(3n + 2)}</math>. So now we only need to find the next value of <math>n</math> that makes <math>\log_5{(3n + 2)}</math> an integer. This means that <math>3n + 2</math> must be a power of <math>5</math>. We test <math>25</math>: <cmath>3n + 2 = 25</cmath> <cmath>3n = 23</cmath> This has no integral solutions, so we try <math>125</math>: <cmath>3n + 2 = 125</cmath> <cmath>3n = 123</cmath> <cmath>n = \boxed{041}</cmath> |
== See also == | == See also == |
Latest revision as of 13:14, 24 November 2020
Problem
The sequence satisfies and for . Let be the least integer greater than for which is an integer. Find .
Solution
The best way to solve this problem is to get the iterated part out of the exponent: Plug in to see the first few terms of the sequence: We notice that the terms are in arithmetic progression. Since , we can easily use induction to show that . So now we only need to find the next value of that makes an integer. This means that must be a power of . We test : This has no integral solutions, so we try :
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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