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

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== Problem ==
 
== Problem ==
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== Solution ==
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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 1==
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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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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=2^{m+1}-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>45</math> and <math>109</math>, that is, <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 ==
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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} - N</math> is the difference between the next power of 2 above <math> 2^{\lfloor\log_{2}(N-1)\rfloor}</math> and <math>N</math>. 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>
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== Solution 3 (using the answer choices) ==
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Note that each <math>N</math> is <math>19</math> less than a power of <math>2</math>. So, the answer will be <math>38</math> less than the sum of <math>2</math> powers of <math>2</math>. Adding <math>38</math> to each answer, we get <math>76</math>, <math>128</math>, <math>192</math>, <math>444</math>, and <math>1062</math>. Obviously we can take out <math>76</math> and <math>1062</math>. Also, <math>128</math> will not work because two powers of two will never sum to another power of <math>2</math> (unless they are equal, which is a contradiction to the question). So, we have <math>192</math> and <math>444</math>. Note that <math>444 = 1 + 443 = 2 + 442 = 4 + 440 = 8 + 436 = 16 + 428 = 32 + 412</math>, etc. We quickly see that <math>444</math> will not work, leaving <math>192</math> which corresponds to <math>\boxed{\textbf{C}}</math>. We can also confirm that this works because <math>192 = 128 + 64 = 2^7 + 2^6</math>.
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== Solution 4 (removing the log) ==
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In order to fix the exponent and get rid of the logarithm term, let <math>N = 2^m + k + 1</math>, with <math>0 \leq k < 2^m</math>. Doing so, we see that <math>\lfloor \log_2{N - 1} \rfloor = m</math>, which turns our given relation into
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<cmath>2^m = 20 + k,</cmath>
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for which the solutions of the form <math>(m, k)</math>, <math>(5, 12)</math> and <math>(6, 44)</math>, follow trivially. Adding up the two values of <math>N</math> gives us <math>32 + 12 + 1 + 64 + 44 + 1 = 154</math>, so the answer is <math>\boxed{\textbf{C}}</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}}
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{{MAA Notice}}

Latest revision as of 10:12, 3 December 2023

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 $45$ and $109$, that is, $2^6-19$ 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 $2^{1+\lfloor\log_{2}(N-1)\rfloor} - N$ is the difference between the next power of 2 above $2^{\lfloor\log_{2}(N-1)\rfloor}$ and $N$. 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 $45 + 109 = 154 \rightarrow \boxed{\textbf{C}}$



Solution 3 (using the answer choices)

Note that each $N$ is $19$ less than a power of $2$. So, the answer will be $38$ less than the sum of $2$ powers of $2$. Adding $38$ to each answer, we get $76$, $128$, $192$, $444$, and $1062$. Obviously we can take out $76$ and $1062$. Also, $128$ will not work because two powers of two will never sum to another power of $2$ (unless they are equal, which is a contradiction to the question). So, we have $192$ and $444$. Note that $444 = 1 + 443 = 2 + 442 = 4 + 440 = 8 + 436 = 16 + 428 = 32 + 412$, etc. We quickly see that $444$ will not work, leaving $192$ which corresponds to $\boxed{\textbf{C}}$. We can also confirm that this works because $192 = 128 + 64 = 2^7 + 2^6$.

Solution 4 (removing the log)

In order to fix the exponent and get rid of the logarithm term, let $N = 2^m + k + 1$, with $0 \leq k < 2^m$. Doing so, we see that $\lfloor \log_2{N - 1} \rfloor = m$, which turns our given relation into \[2^m = 20 + k,\] for which the solutions of the form $(m, k)$, $(5, 12)$ and $(6, 44)$, follow trivially. Adding up the two values of $N$ gives us $32 + 12 + 1 + 64 + 44 + 1 = 154$, so the answer is $\boxed{\textbf{C}}$.

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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