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

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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 $192 = 128 + 64 = 2^7 + 2^6$ .

@@ Line 14: / Line 14: @@
 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>.
-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>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>
 == Solution 2 ==

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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