1997 AHSME Problems/Problem 21

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Problem

For any positive integer $n$, let

$f(n) =\left\{\begin{matrix}\log_{8}{n}, &\text{if }\log_{8}{n}\text{ is rational,}\\ 0, &\text{otherwise.}\end{matrix}\right.$

What is $\sum_{n = 1}^{1997}{f(n)}$?

$\textbf{(A)}\ \log_{8}{2047}\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ \frac{55}{3}\qquad\textbf{(D)}\ \frac{58}{3}\qquad\textbf{(E)}\ 585$

Solution

For positive integers $n$, $\log_8 n$ is rational if and only if $n = 2^k$ for an integer $k$. That's because $\log_8 2^k = k\log_8 2 = \frac{k}{3}$.

So we actually want to find $\sum_{k=0}^{10} \log_8 2^k$, since $2^{11}$ will be over $1997$.

Using log properties, we get $\sum_{k=0}^{10} k \log_8 2$

$\frac{1}{3}\sum_{k=0}^{10} k$

$\frac{1}{3}\cdot (\frac{10\cdot 11}{2})$

$\frac{55}{3}$, and the answer is $\boxed{C}$

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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