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

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

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

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

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

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 $m$ .

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

$m \le \lfloor\log_{2}(N-1) < m+1$

$2^{m}+1 \le N < 2^{m+1}+1$

$2^{m}+1 \le 2^{m+1}-19 < 2^{m+1}+1$

$2^{m}+20 \le 2^{m+1} < 2^{m+1}+20$

The two smallest possible value of $m$ where $m$ is a positive integers are $5$ and $6$ respectively.

Sum of the two smallest possible value of $m = 2^{5+1}-19+2^{6+1}-19=154$

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2011 AMC 12A Problems/Problem 19

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