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

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== Problem ==
 
== Problem ==
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For each real number <math>x</math>, let <math>\lfloor x \rfloor</math> denote the [[greatest integer function|greatest integer]] that does not exceed x. For how man positive integers <math>n</math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer?
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For each real number <math>x</math>, let <math>\lfloor x \rfloor</math> denote the [[greatest integer function|greatest integer]] that does not exceed x. For how many positive integers <math>n</math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer?
  
 
== Solution ==
 
== Solution ==
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n must satisfy these [[inequality|inequalities]]:
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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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<div style="text-align:center;">
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<math>256\leq n<512</math></div>
 
<math>256\leq n<512</math></div>
  
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There are 4 for the first inequality, 16 for the second, 64 for the third, and 256 for the fourth. <math>4+16+64+256=340</math>
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There are <math>4</math> for the first inequality, <math>16</math> for the second, <math>64</math> for the third, and <math>256</math> for the fourth, so the answer is <math>4+16+64+256=\boxed{340}</math>.
  
 
== See also ==
 
== See also ==
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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 (Problems • Answer Key • Resources)
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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