AMC12 2011 A

AMC12 2011 A · Q19

AMC12 2011 A · Q19. It mainly tests Inequalities with floors/ceilings (basic), Powers & residues.

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 $N$?

在一场比赛中有 $N$ 名选手，获得精英身份的选手数等于 $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$。假设有 $19$ 名选手获得精英身份。$N$ 的两个最小可能值的和是多少？

(A) 38 38

(B) 90 90

(D) 406 406

(E) 1024 1024

Answer

Correct choice: (C)

正确答案：（C）

Solution

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

从 $2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19$ 出发，移项得到 $\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)$。由于 $\lfloor\log_{2}(N-1)\rfloor$ 是正整数，$\frac{N+19}{2}$ 必须形如 $2^{m}$（其中 $m$ 为正整数）。因此 $N=2^{m+1}-19$。检验满足条件的整数 $N$（从 $N=19$ 开始）可得最小的两个可行值为 $45$ 和 $109$，即 $2^6-19$ 与 $2^7-19$，对应 $m$ 分别为 $5$ 与 $6$。两者之和为 $45 + 109 = 154 \rightarrow \boxed{\textbf{C}}$。

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