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Difference between revisions of "2014 AIME II Problems/Problem 7"

(Solution)
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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 possible values of <math>n=3</math> and <math>n=18</math> so our desired answer is <math>3+18=\boxed{021}</math>
+
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 $f(x)=(x^2+3x+2)^{\cos(\pi x)}$. Find the sum of all positive integers $n$ for which \[\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.\]

Solution

Note that $\cos(\pi x)$ is $-1$ when $x$ is odd and $1$ when $x$ is even. Also note that $x^2+3x+2=(x+1)(x+2)$ for all $x$. Therefore \[\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if x is even}\] \[\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if x is odd}\] Because of this, $\sum_{k=1}^n\log_{10}f(k)$ is a telescoping series of logs, and we have \[\sum_{k=1}^n\log_{10}f(k)= \log_{10}(n+2)-\log_{10}2=\log_{10}\frac{n+2}{2}\ \ \ \text{if n is even}\] \[\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if n is odd}\] Setting each of the above quantities to $1$ and $-1$ and solving for $n$, we get integer values of $n=3$ and $n=18$ so our desired answer is $3+18=\boxed{021}$

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