# 1997 AHSME Problems/Problem 21

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS