Difference between revisions of "2013 AIME II Problems/Problem 2"

Revision as of 16:23, 19 September 2022

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 argument 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 ${(a^x)^{y}}=a^{xy}$ , we simplify to $2^{b\cdot2^{a}}=2^{1000}$ . Eliminating equal bases leaves $b\cdot2^a=1000$ . 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 $\boxed{881}$ .

Note that $a$ cannot be $0,$ since that would cause the $\log_{2^a}$ to have a $1$ in the base, which is not possible.

Video Solution

https://youtu.be/zf9ld5KL_g4

~Lucas

@@ Line 11: / Line 11: @@
 ==Video Solution==
 https://youtu.be/zf9ld5KL_g4
 ~Lucas

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Difference between revisions of "2013 AIME II Problems/Problem 2"

Revision as of 16:23, 19 September 2022

Contents

Problem 2

Solution

Video Solution

See also