Difference between revisions of "1996 AIME Problems/Problem 2"

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== Solution ==
 
== Solution ==
For integers <math>k</math>, we want <math>\lfloor \log_2 n\rfloor = 2k</math>, or <math>2k \le \log_2 < 2k+1 \Longrightarrow 2^{2k} \le n < 2^{2k+1}</math>. Thus, <math>n</math> must satisfy these [[inequality|inequalities]] (since <math>n < 1000</math>):
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For integers <math>k</math>, we want <math>\lfloor \log_2 n\rfloor = 2k</math>, or <math>2k \le \log_2 n < 2k+1 \Longrightarrow 2^{2k} \le n < 2^{2k+1}</math>. Thus, <math>n</math> must satisfy these [[inequality|inequalities]] (since <math>n < 1000</math>):
  
 
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 18:31, 4 July 2013

Problem

For each real number $x$, let $\lfloor x \rfloor$ denote the greatest integer that does not exceed x. For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_{2} n \rfloor$ is a positive even integer?

Solution

For integers $k$, we want $\lfloor \log_2 n\rfloor = 2k$, or $2k \le \log_2 n < 2k+1 \Longrightarrow 2^{2k} \le n < 2^{2k+1}$. Thus, $n$ must satisfy these inequalities (since $n < 1000$):

$4\leq n <8$
$16\leq n<32$
$64\leq n<128$

$256\leq n<512$

There are $4$ for the first inequality, $16$ for the second, $64$ for the third, and $256$ for the fourth, so the answer is $4+16+64+256=\boxed{340}$.

See also

1996 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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