Difference between revisions of "1997 AHSME Problems/Problem 21"
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<math> \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 </math> | <math> \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 </math> | ||
| + | ==Solution== | ||
| + | |||
| + | For positive integers <math>n</math>, <math>\log_8 n</math> is rational if and only if <math>n = 2^k</math> for an integer <math>k</math>. That's because <math>\log_8 2^k = k\log_8 2 = \frac{k}{3}</math>. | ||
| + | |||
| + | So we actually want to find <math>\sum_{k=0}^{10} \log_8 2^k</math>, since <math>2^{11}</math> will be over <math>1997</math>. | ||
| + | |||
| + | Using log properties, we get <math>\sum_{k=0}^{10} k \log_8 2</math> | ||
| + | |||
| + | <math>\frac{1}{3}\sum_{k=0}^{10} k</math> | ||
| + | |||
| + | <math>\frac{1}{3}\cdot (\frac{10\cdot 11}{2})</math> | ||
| + | |||
| + | <math>\frac{55}{3}</math>, and the answer is <math>\boxed{C}</math> | ||
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=20|num-a=22}} | {{AHSME box|year=1997|num-b=20|num-a=22}} | ||
Revision as of 18:06, 9 August 2011
Problem
For any positive integer 
, let
What is 
?
Solution
For positive integers , 
is rational if and only if 
for an integer 
. That's because 
.
So we actually want to find 
, since 
will be over 
.
Using log properties, we get 
, and the answer is 
See also
| 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 | ||
