2013 AIME II Problems/Problem 2

Positive integers $a$ and $b$ satisfy the condition $\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.$ Find the sum of all possible values of $a+b$ .

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 arguement the result. That means $\log_{2^a}(\log_{2^b}(2^{1000}))=1$ (Because $2^0=1$ ). Doing this again, we get $\log_{2^b}(2^{1000})=2^a$ . Doing the process one more time, we finally eliminate all of the logs, getting ${2^{b}}^{2^a}=2^{1000}$ . Using the property that ${a^{x^{y}}=a^{xy}$ (Error compiling LaTeX. Unknown error_msg), we simplify to $2^{b*2^{a}}=2^{1000}$ . Eliminating equal bases leaves $b*2^a=1000$ . The largest a such that $2^a$ divides $1000$ is $3$ , so we only need to check $1$ , $2$ , and $3$ . When $a=1$ , $b=500$ ; when $a=2$ , $b=250$ ; when $a=3$ , $b=125$ . Summing all the a's and b's gives the answer of $\boxed{881}$

Page

Toolbox

Search

2013 AIME II Problems/Problem 2

Solution