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

(New page: == Problem == The sequence <math>(a_n)</math> satisfies <math>a_1 = 1</math> and <math>5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}</math> for <math>n \geq 1</math>. Let <math>k...)
 
(Solution)
Line 9: Line 9:
 
<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}</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{41}</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{41}</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 15:15, 20 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 $125$: \[3n + 2 = 125\] \[3n = 123\] \[n = \boxed{41}\]

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