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

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== Solution ==
 
== Solution ==
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<math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19 </math>
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<math> 2^{1+\lfloor\log_{2}(N-1)\rfloor} = N+19 </math>
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<math> 1+\lfloor\log_{2}(N-1)\rfloor = \log_{2}(N+19) </math>
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<math> \lfloor\log_{2}(N-1)\rfloor = \log_{2} (\frac{N+19}{2}) </math>
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Since <math> \lfloor\log_{2}(N-1)\rfloor </math> is a positive integer, <math> \frac{N+19}{2}</math> must be in the form of <math>2^{m} </math> for some positive integer <math> m </math>.
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<math> N=2^{m+1}-19 </math>
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<math> m \le \lfloor\log_{2}(N-1) < m+1 </math>
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<math> 2^{m}+1 \le N < 2^{m+1}+1 </math>
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<math> 2^{m}+1 \le 2^{m+1}-19 < 2^{m+1}+1 </math>
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<math> 2^{m}+20 \le 2^{m+1} < 2^{m+1}+20 </math>
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The two smallest possible value of <math> m </math> where <math> m </math> is a positive integers are <math> 5 </math> and <math> 6 </math> respectively.
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Sum of the two smallest possible value of <math> m = 2^{5+1}-19+2^{6+1}-19=154 </math>
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== 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 03:03, 11 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

$2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19$


$2^{1+\lfloor\log_{2}(N-1)\rfloor} = N+19$


$1+\lfloor\log_{2}(N-1)\rfloor = \log_{2}(N+19)$


$\lfloor\log_{2}(N-1)\rfloor = \log_{2} (\frac{N+19}{2})$


Since $\lfloor\log_{2}(N-1)\rfloor$ is a positive integer, $\frac{N+19}{2}$ must be in the form of $2^{m}$ for some positive integer $m$.


$N=2^{m+1}-19$


$m \le \lfloor\log_{2}(N-1) < m+1$


$2^{m}+1 \le N < 2^{m+1}+1$


$2^{m}+1 \le 2^{m+1}-19 < 2^{m+1}+1$


$2^{m}+20 \le 2^{m+1} < 2^{m+1}+20$

The two smallest possible value of $m$ where $m$ is a positive integers are $5$ and $6$ respectively.

Sum of the two smallest possible value of $m = 2^{5+1}-19+2^{6+1}-19=154$

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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