Difference between revisions of "2011 AMC 12A Problems/Problem 19"

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== Problem ==
 
== Problem ==
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At a competition with <math>N</math> players, the number of players given elite status is equal to <math>2^{1+\lfloor \log_{2} (N-1) \rfloor}-N</math>. Suppose that <math>19</math> players are given elite status. What is the sum of the two smallest possible values of <math>N</math>?
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<math>
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\textbf{(A)}\ 38 \qquad
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\textbf{(B)}\ 90 \qquad
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\textbf{(C)}\ 154 \qquad
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\textbf{(D)}\ 406 \qquad
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\textbf{(E)}\ 1024 </math>
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== Solution ==
 
== Solution ==
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=18|num-a=20|ab=A}}
 
{{AMC12 box|year=2011|num-b=18|num-a=20|ab=A}}

Revision as of 01:36, 10 February 2011

Problem

At a competition with $N$ players, the number of players given elite status is equal to $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$. Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$?

$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 154 \qquad \textbf{(D)}\ 406 \qquad \textbf{(E)}\ 1024$

Solution

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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
All AMC 12 Problems and Solutions
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