AMC10 2010 B

AMC10 2010 B · Q21

AMC10 2010 B · Q21. It mainly tests Divisibility & factors, Remainders & modular arithmetic.

A palindrome between 1000 and 10,000 is chosen at random. What is the probability that it is divisible by 7?

在1000到10,000之间随机选择一个回文数。它能被7整除的概率是多少？

(A) 1/10 1/10

(B) 1/9 1/9

(D) 1/6 1/6

(E) 1/5 1/5

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Each four-digit palindrome has digit representation $abba$ with $1 \le a \le 9$ and $0 \le b \le 9$. The value of the palindrome is $1001a + 110b$. Because $1001$ is divisible by $7$ and $110$ is not, the palindrome is divisible by $7$ if and only if $b = 0$ or $b = 7$. Thus the requested probability is $\frac{2}{10} = \frac{1}{5}$.

答案（E）：每个四位回文数都可表示为数字形式 $abba$，其中 $1 \le a \le 9$ 且 $0 \le b \le 9$。该回文数的值为 $1001a + 110b$。由于 $1001$ 能被 $7$ 整除而 $110$ 不能，所以该回文数当且仅当 $b=0$ 或 $b=7$ 时能被 $7$ 整除。因此所求概率为 $\frac{2}{10}=\frac{1}{5}$。

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