AMC12 2006 B

AMC12 2006 B · Q20

AMC12 2006 B · Q20. It mainly tests Logarithms (rare), Inequalities with floors/ceilings (basic).

Let $x$ be chosen at random from the interval $(0,1)$. What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$? Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.

从区间 $(0,1)$ 中随机选择 $x$。满足 $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ 的概率是多少？这里 $\lfloor x\rfloor$ 表示不超过 $x$ 的最大整数。

(A) $\frac{1}{8}$ $\frac{1}{8}$

(B) $\frac{3}{20}$ $\frac{3}{20}$

(D) $\frac{1}{5}$ $\frac{1}{5}$

(E) $\frac{1}{4}$ $\frac{1}{4}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $k$ be an arbitrary integer. For which $x$ do we have $\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k$? The equation $\lfloor\log_{10}x\rfloor = k$ can be rewritten as $10^k \leq x < 10^{k+1}$. The second one gives us $10^k \leq 4x < 10^{k+1}$. Combining these, we get that both hold at the same time if and only if $10^k \leq x < \frac{10^{k+1}}4$. Hence for each integer $k$ we get an interval of values for which $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$. These intervals are obviously pairwise disjoint. For any $k\geq 0$ the corresponding interval is disjoint with $(0,1)$, so it does not contribute to our answer. On the other hand, for any $k<0$ the entire interval is inside $(0,1)$. Hence our answer is the sum of the lengths of the intervals for $k<0$. For a fixed $k$ the length of the interval $\left[ 10^k, \frac{10^{k+1}}4 \right)$ is $\frac 32\cdot 10^k$. This means that our result is $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{\frac 16}$.

令 $k$ 为任意整数。对于哪些 $x$ 有 $\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k$？方程 $\lfloor\log_{10}x\rfloor = k$ 可改写为 $10^k \leq x < 10^{k+1}$。另一个给出 $10^k \leq 4x < 10^{k+1}$。合并可得，两者同时成立当且仅当 $10^k \leq x < \frac{10^{k+1}}4$。因此对每个整数 $k$，我们得到一个使得 $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ 的区间。这些区间显然两两不相交。当 $k\geq 0$ 时，对应区间与 $(0,1)$ 不相交，因此对答案无贡献。另一方面，当 $k<0$ 时，整个区间都在 $(0,1)$ 内。因此答案是所有 $k<0$ 的这些区间长度之和。对固定的 $k$，区间 $\left[ 10^k, \frac{10^{k+1}}4 \right)$ 的长度为 $\frac 32\cdot 10^k$。因此结果为 $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{\frac 16}$。

Topics