2013 AIME II Problems/Problem 2

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 1

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 (also the problem specifies that $a$ and $b$ are positive).

Solution 2

We proceed as in Solution 1, raising $2$ to both sides to achieve $\log_{2^a}(\log_{2^b}(2^{1000})) = 1.$ We raise $2^a$ to both sides to get $\log_{2^b}(2^{1000})=2^a$, then simplify to get $\dfrac{1000}b=2^a$.

At this point, we want both $a$ and $b$ to be integers. Thus, $2^a$ can only be a power of $2$. To help us see the next step, we factorize $1000$: $\dfrac{2^35^3}b=2^a.$ It should be clear that $a$ must be from $1$ to $3$; when $a=1$, $b=500$; when $a=2$, $b=250$; and finally, when $a=3$, $b=125.$ We sum all the pairs to get $\boxed{881}.$


Video Solution



See also

2013 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions

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