AMC12 2002 B

AMC12 2002 B · Q15

AMC12 2002 B · Q15. It mainly tests Fractions, Divisibility & factors.

How many four-digit numbers $N$ have the property that the three-digit number obtained by removing the leftmost digit is one ninth of $N$?

有多少个四位数 $N$ 具有这样的性质：去掉最左边数字得到的三位数是 $N$ 的九分之一？

(A) 4 4

(B) 5 5

(D) 7 7

(E) 8 8

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $N = \overline{abcd} = 1000a + \overline{bcd}$, such that $\frac{N}{9} = \overline{bcd}$. Then $1000a + \overline{bcd} = 9\overline{bcd} \Longrightarrow 125a = \overline{bcd}$. Since $100 \le \overline{bcd} < 1000$, from $a = 1, \ldots, 7$ we have $7$ three-digit solutions, and the answer is $\mathrm{(D)}$.

设 $N = \overline{abcd} = 1000a + \overline{bcd}$，且 $\frac{N}{9} = \overline{bcd}$。则 $1000a + \overline{bcd} = 9\overline{bcd} \Longrightarrow 125a = \overline{bcd}$。由于 $100 \le \overline{bcd} < 1000$，当 $a = 1, \ldots, 7$ 时共有 $7$ 个三位数解，因此答案是 $\mathrm{(D)}$。

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