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 01:13, 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:
![\[5^{(a_{n + 1} - a_n)} = \frac {1}{n + \frac {2}{3}} + 1\]](http://latex.artofproblemsolving.com/8/3/d/83dead807fbda0d002949edbcc2c4dc658cc3cc3.png)
![\[5^{(a_{n + 1} - a_n)} = \frac {n + \frac {5}{3}}{n + \frac {2}{3}}\]](http://latex.artofproblemsolving.com/b/1/2/b12143dfb49fc4d354f9f7d04c207e216e273598.png)
![\[5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}\]](http://latex.artofproblemsolving.com/f/2/c/f2c016a16c6306dd03125b535bb2a994f21a6c7c.png)
![\[a_{n + 1} - a_n = \log_5{\frac {3n + 5}{3n + 2}}\]](http://latex.artofproblemsolving.com/b/0/c/b0c12ebb34f6ce5924bf769b0c5d2b84bafb3159.png)
![\[a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}\]](http://latex.artofproblemsolving.com/4/9/d/49d4c2ec3f3c49ea4bcc502c25f6002485d4b4fa.png)
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 
: ![\[3n + 2 = 25\]](http://latex.artofproblemsolving.com/c/d/2/cd2c214fe355cff53f5df66998eff2a8c507be8f.png)
![\[3n = 22\]](http://latex.artofproblemsolving.com/a/8/6/a869731f2e1dd11ca0f49867e53e5023952563d6.png)
This has no integral solutions, so we try 
: ![\[3n + 2 = 125\]](http://latex.artofproblemsolving.com/f/2/c/f2cd83ce3666d32d3f20543e9c1a055a2e7a4103.png)
![\[3n = 123\]](http://latex.artofproblemsolving.com/9/3/5/9358b205bc6c0a18e0a042adc3ad937a1d02a4e8.png)
![\[n = \boxed{041}\]](http://latex.artofproblemsolving.com/c/b/e/cbe61593c6c18016e2ee0a442eff90becb2e62a0.png)
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 | ||
