AMC12 2004 B

AMC12 2004 B · Q25

AMC12 2004 B · Q25. It mainly tests Patterns & sequences (misc), Powers & residues.

Given that $2^{2004}$ is a $604$-digit number whose first digit is $1$, how many elements of the set $S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}$ have a first digit of $4$?

已知 $2^{2004}$ 是一个 $604$ 位数，且其首位数字为 $1$。问集合 $S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}$ 中有多少个元素的首位数字为 $4$？

(A) 194 194

(B) 195 195

(D) 197 197

(E) 198 198

Answer

Correct choice: (B)

正确答案：（B）

Solution

Given $n$ digits, there must be exactly one power of $2$ with $n$ digits such that the first digit is $1$. Thus $S$ contains $603$ elements with a first digit of $1$. For each number in the form of $2^k$ such that its first digit is $1$, then $2^{k+1}$ must either have a first digit of $2$ or $3$, and $2^{k+2}$ must have a first digit of $4,5,6,7$. Thus there are also $603$ numbers with first digit $\{2,3\}$ and $603$ numbers with first digit $\{4,5,6,7\}$. By using complementary counting, there are $2004 - 3 \times 603 = 195$ elements of $S$ with a first digit of $\{8,9\}$. Now, $2^k$ has a first digit of $\{8,9\}$ if and only if the first digit of $2^{k-1}$ is $4$, so there are $\boxed{195} \Rightarrow \mathrm{(B)}$ elements of $S$ with a first digit of $4$.

对于任意位数为 $n$ 的数，必恰有一个 $2$ 的幂是 $n$ 位数且首位数字为 $1$。因此 $S$ 中有 $603$ 个元素的首位数字为 $1$。对每个形如 $2^k$ 且首位数字为 $1$ 的数，$2^{k+1}$ 的首位数字必为 $2$ 或 $3$，而 $2^{k+2}$ 的首位数字必为 $4,5,6,7$ 之一。因此首位数字为 $\{2,3\}$ 的数也有 $603$ 个，首位数字为 $\{4,5,6,7\}$ 的数也有 $603$ 个。用补集计数，$S$ 中首位数字为 $\{8,9\}$ 的元素个数为 $2004 - 3 \times 603 = 195$。注意到 $2^k$ 的首位数字为 $\{8,9\}$ 当且仅当 $2^{k-1}$ 的首位数字为 $4$，因此 $S$ 中首位数字为 $4$ 的元素个数为 $\boxed{195} \Rightarrow \mathrm{(B)}$。

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