Question

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

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{7}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

答案：E

知识点：回文数整除性基本概率

难度评级：1540

解答：

任意一个 $4$ 位数都可写成 $abcd$ ，展开后为对回文数， $a = d$ 且 $b = c$ ，所以可化为因为 $1001$ 能被 $7$ 整除，所以还需要 $110b$ 能被 $7$ 整除。 $10^3a + 10^2b + 10c + d.$ $1001a + 110b.$

$110$ 不能被 $7$ 整除，因此 $b$ 必须为 $0$ 或 $7$ 。

有 $9$ 种 $a$ 和 $2$ 种 $b$ ，符合条件的回文数共有 $9 \cdot 2 = 18$ 个。

四位回文数总数为 $9 \cdot 10$ ，因为千位有 $9$ 种选择、百位有 $10$ 种选择。

所求概率为 $\dfrac{18}{9 \cdot 10} = \dfrac{1}{5}.$

所以正确答案是 E。

Note that we can express any $4$ digit number as $abcd.$ This can be expressed in long form as $10^3a + 10^2b + 10c + d.$ Since in a palindrome, we have that $a = d$ and $b = c.$ We can simplify this to get $1001a + 110b.$ Note that $1001$ is divisible by $7.$ This means that $110b$ must also be divisible by $7.$

The only way for this to happen is if $b$ is $0$ or $7$ since $110$ is not divisible by $7.$

There are $9$ options for $a$ and $2$ options for $b,$ for a total of $9 \cdot 2 = 18$ palindromes.

The total number of palindromes is $9 \cdot 10$ since there are $9$ options for the thousands digit and $10$ options for the hundreds digit.

The desired probability is then $\dfrac{18}{9 \cdot 10} = \dfrac{1}{5}.$

Thus, E is the correct answer.

2010 AMC 10B 第 21 题

答案：E

难度评级：1540

解答：

其他年份的第 21 题