Difference between revisions of "2014 AIME II Problems/Problem 7"
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Latest revision as of 19:13, 2 December 2017
Let . Find the sum of all positive integers for which
First, let's split it into two cases to get rid of the absolute value sign
Now we simplify using product-sum logarithmic identites:
Note that the exponent is either if is odd or if is even.
Writing out the first terms we have
This product clearly telescopes (i.e. most terms cancel) and equals either or . But the resulting term after telescoping depends on parity (odd/evenness), so we split it two cases, one where is odd and another where is even.
For odd , it telescopes to where is clearly .
For even , it telescopes to where is the only possible value. Thus the answer is
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 possible values of and so our desired answer is
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