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

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\textbf{(E)}\ 1024 </math>
 
\textbf{(E)}\ 1024 </math>
  
== Solution ==
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== Solution 1==
 
We start with <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19</math>. After rearranging, we get <math>\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)</math>.
 
We start with <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19</math>. After rearranging, we get <math>\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)</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>
 
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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 +
== Solution 2==
 +
We examine the value that <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}</math> takes over various intervals. The <math>\lfloor\log_{2}(N-1)\rfloor</math> means it changes on each multiple of 2, like so:
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 +
2 --> 1
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 +
3 - 4 --> 2
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 +
5 - 8 --> 3
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9 - 16 --> 4
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From this, we see that <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}</math> is the difference between <math>N</math> and the next power of 2 above it. We are looking for <math>N</math> such that this difference is 19. The first two <math>N</math> that satisfy this are <math>45 = 64-19</math> and <math>109=128-19</math> for a final answer of <math>45 + 109 = 154 \rightarrow \boxed{\textbf{C}}</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:45, 9 September 2017

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 1

We start with $2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19$. After rearranging, we get $\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)$.

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

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 $45 + 109 = 154 \rightarrow \boxed{\textbf{C}}$

Solution 2

We examine the value that $2^{1+\lfloor\log_{2}(N-1)\rfloor}$ takes over various intervals. The $\lfloor\log_{2}(N-1)\rfloor$ means it changes on each multiple of 2, like so:

2 --> 1

3 - 4 --> 2

5 - 8 --> 3

9 - 16 --> 4

From this, we see that $2^{1+\lfloor\log_{2}(N-1)\rfloor}$ is the difference between $N$ and the next power of 2 above it. We are looking for $N$ such that this difference is 19. The first two $N$ that satisfy this are $45 = 64-19$ and $109=128-19$ for a final answer of $45 + 109 = 154 \rightarrow \boxed{\textbf{C}}$

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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