2013 AIME II Problems/Problem 2
Problem 2
Positive integers and satisfy the condition Find the sum of all possible values of .
Solution
To simplify, we write this logarithmic expression as an exponential one. Just looking at the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means (because ). Doing this again, we get . Doing the process one more time, we finally eliminate all of the logs, getting . Using the property that , we simplify to . Eliminating equal bases leaves . The largest such that divides is , so we only need to check ,, and . When , ; when , ; when , . Summing all the 's and 's gives the answer of .
Note that cannot be since that would cause the to have a in the base, which is not possible.
See also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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