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

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== Solution ==
 
== Solution ==
 +
We start with <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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After rearranging, we get
  
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<math>\lfloor\log_{2}(N-1)\rfloor = \log_{2} (\frac{N+19}{2})</math>.
  
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<math> 2^{1+\lfloor\log_{2}(N-1)\rfloor} = N+19 </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>. From this fact, we get <math>N+19=2^{m+1}</math>.
  
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<math> 1+\lfloor\log_{2}(N-1)\rfloor = \log_{2}(N+19) </math>
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If we now check integer values of N that satisfy this condition, starting from <math>N=19</math>, we quickly see that the first values that work for <math>N</math> are <math>2^6 -19</math> and <math>2^7 -19</math>, giving values of <math>5</math> and <math>6</math> for <math>m</math>, respectively. Adding up these two values for <math>N</math>, we get <math>45 + 109 = 154 \rightarrow \boxed{\textbf{C}}</math>
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<math> \lfloor\log_{2}(N-1)\rfloor = \log_{2} (\frac{N+19}{2}) </math>
 
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−
 
 
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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>
 
  
 
== 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 21:29, 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

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

After rearranging, we get

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


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 $m$. From this fact, we get $N+19=2^{m+1}$.

If we now check integer values of N that satisfy this condition, starting from $N=19$, we quickly see that the first values that work for $N$ are $2^6 -19$ and $2^7 -19$, giving values of $5$ and $6$ for $m$, respectively. Adding up these two values for $N$, we get $45 + 109 = 154 \rightarrow \boxed{\textbf{C}}$

See also

2011 AMC 12A (Problems • Answer Key • Resources)
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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