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

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<math>4+16+64+256=340</math>
 
<math>4+16+64+256=340</math>
 
== See also ==
 
== See also ==
* [[1996 AIME Problems/Problem 1 | Previous problem]]
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*[[1996 AIME Problems]]
* [[1996 AIME Problems/Problem 3 | Next problem]]
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* [[1996 AIME Problems]]
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{{AIME box|year=1996|num-b=1|num-a=3}}

Revision as of 15:57, 24 September 2007

Problem

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

Solution

n must satisfy these inequalities:


$4\leq n <8$

$16\leq n<32$

$64\leq n<128$

$256\leq n<512$

There are 4 for the first inequality, 16 for the second, 64 for the third, and 256 for the fourth.

$4+16+64+256=340$

See also

1996 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions