# 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

## Solution

Note that is when is odd and when is even. Also note that for all . Therefore Because of this, is a telescoping series of logs, and we have Setting each of the above quantities to and and solving for , we get integer values of and so our desired answer is