2009 AIME I Problems/Problem 7: Difference between revisions
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We can now test powers of <math>5</math>. | We can now test powers of <math>5</math>. | ||
<math>5</math>-that gives us <math>n=1</math>, which is useless. | <math>5</math> - that gives us <math>n=1</math>, which is useless. | ||
<math>25</math>-that gives a non-integer <math>n</math>. | <math>25</math> - that gives a non-integer <math>n</math>. | ||
<math>125</math>-that | <math>125</math> - that gives <math>n=\boxed{41}</math>. | ||
-integralarefun | -integralarefun | ||
Revision as of 19:51, 30 November 2022
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)
![\[a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}\]](http://latex.artofproblemsolving.com/1/f/7/1f7b9dd9bc69ad1dacbe9d6bda96cec3c1ef7c68.png)
![\[a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}\]](http://latex.artofproblemsolving.com/4/9/d/49d4c2ec3f3c49ea4bcc502c25f6002485d4b4fa.png)
Plug in 
to see the first few terms of the sequence: ![\[\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.\]](http://latex.artofproblemsolving.com/4/a/8/4a89100b34f9f3e24be5147ec8db483a211f6c1b.png)
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 
: ![\[3n + 2 = 25\]](http://latex.artofproblemsolving.com/c/d/2/cd2c214fe355cff53f5df66998eff2a8c507be8f.png)
![\[3n = 23\]](http://latex.artofproblemsolving.com/d/f/9/df919830eba0bb5ea8b65456fbbf31df25a727b8.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)
Solution 2 (Telescoping?)
We notice that by multiplying the equation from an arbitrary 
all the way to 
, we get:
![\[5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}\]](http://latex.artofproblemsolving.com/1/b/0/1b0b996ee2fb8341f322887efd31c5382e543825.png)
This simplifies to
![\[5^{a_n}=3n+2.\]](http://latex.artofproblemsolving.com/f/8/e/f8e5574565bc12e4564a61a1397568b80d3dd245.png)
We can now test powers of .
- that gives us 
, which is useless.
- that gives a non-integer .
- that gives 
.
-integralarefun
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
