Difference between revisions of "2009 AIME I Problems/Problem 7"

(Solution)
m (Solution)
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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 $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$. Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$.

Solution

The best way to solve this problem is to get the iterated part out of the exponent: \[5^{(a_{n + 1} - a_n)} = \frac {1}{n + \frac {2}{3}} + 1\] \[5^{(a_{n + 1} - a_n)} = \frac {n + \frac {5}{3}}{n + \frac {2}{3}}\] \[5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}\] \[a_{n + 1} - a_n = \log_5{\frac {3n + 5}{3n + 2}}\] \[a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}\] Since $a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}$, we can easily use induction to show that $a_n = \log_5{(3n + 2)}$. So now we only need to find the next value of $n$ that makes $\log_5{(3n + 2)}$ an integer. This means that $3n + 2$ must be a power of $5$. We test $25$: \[3n + 2 = 25\] \[3n = 22\] This has no integral solutions, so we try $125$: \[3n + 2 = 125\] \[3n = 123\] \[n = \boxed{041}\]

See also

2009 AIME I (ProblemsAnswer KeyResources)
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