Difference between revisions of "1997 AHSME Problems/Problem 21"

Latest revision as of 13:13, 5 July 2013

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 $\boxed{C}$

1997 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 == See also ==
 {{AHSME box|year=1997|num-b=20|num-a=22}}
+{{MAA Notice}}

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Difference between revisions of "1997 AHSME Problems/Problem 21"

Latest revision as of 13:13, 5 July 2013

Problem

Solution

See also