2011 AMC 12A Problems/Problem 19

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 $\boxed{\textbf{C}}$ .

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