Difference between revisions of "2009 AIME I Problems/Problem 7"
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<cmath>5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}</cmath> | <cmath>5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}</cmath> | ||
<cmath>a_{n + 1} - a_n = \log_5{\frac {3n + 5}{3n + 2}}</cmath> | <cmath>a_{n + 1} - a_n = \log_5{\frac {3n + 5}{3n + 2}}</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 = 22</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> | 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 = 22</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 == | ||
{{AIME box|year=2009|n=I|num-b=6|num-a=8}} | {{AIME box|year=2009|n=I|num-b=6|num-a=8}} |
Revision as of 02:13, 30 March 2009
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:
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 |