Difference between revisions of "2014 AIME II Problems/Problem 7"
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<cmath>\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if n is odd}</cmath> | <cmath>\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if n is odd}</cmath> | ||
Setting each of the above quantities to <math>1</math> and <math>-1</math> and solving for <math>n</math>, | Setting each of the above quantities to <math>1</math> and <math>-1</math> and solving for <math>n</math>, | ||
| − | we get | + | we get integer values of <math>n=3</math> and <math>n=18</math> so our desired answer is <math>3+18=\boxed{021}</math> |
Revision as of 15:58, 27 March 2014
Problem
Let 
. Find the sum of all positive integers 
for which
![\[\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.\]](http://latex.artofproblemsolving.com/3/8/9/389a540c1dec657ab2e68535a6098d77f575cdcb.png)
Solution
Note that 
is 
when 
is odd and 
when is even. Also note that 
for all . Therefore
![\[\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if x is even}\]](http://latex.artofproblemsolving.com/4/0/f/40f85e8eba9571aa627ba0f640b5ff19dd32f565.png)
![\[\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if x is odd}\]](http://latex.artofproblemsolving.com/d/0/f/d0f11dcac165ab35d31544c65b41015f9dd3b486.png)
Because of this, 
is a telescoping series of logs, and we have
![\[\sum_{k=1}^n\log_{10}f(k)= \log_{10}(n+2)-\log_{10}2=\log_{10}\frac{n+2}{2}\ \ \ \text{if n is even}\]](http://latex.artofproblemsolving.com/0/3/c/03cd1e25260c4ede8237be703bd565b956dc7928.png)
![\[\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if n is odd}\]](http://latex.artofproblemsolving.com/b/3/2/b3297f62de2d50a5a4dbdc963e4f91eec2d17771.png)
Setting each of the above quantities to and and solving for ,
we get integer values of 
and 
so our desired answer is 
