Difference between revisions of "1997 AHSME Problems/Problem 21"
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− | ==Problem | + | ==Problem== |
For any positive integer <math>n</math>, let | For any positive integer <math>n</math>, let | ||
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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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:13, 5 July 2013
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 | |
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All AHSME Problems and Solutions |
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