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 16:58, 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