Question

从 $1000$ 到 $10{,}000$ 之间的回文数中随机选一个。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 $7?$

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{7}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

四位回文数形如 $\overline{abba}=1001a+110b$ ，其中 $1\le a\le9$ ， $0\le b\le9$ 。

因为 $1001=7\cdot11\cdot13$ 可被 $7$ 整除，而 $110$ 不能，所以该数能被 $7$ 整除当且仅当 $7\mid b$ ，即 $b=0$ 或 $b=7$ 。

对每个 $a$ ，这是 $b$ 的 $10$ 个选择中的 $2$ 个，概率为 $\dfrac{2}{10}=\dfrac15$ 。

因此，正确答案是 E。

A four-digit palindrome has the form $\overline{abba}=1001a+110b$ with $1\le a\le9$ and $0\le b\le9.$

Since $1001=7\cdot11\cdot13$ is divisible by $7$ and $110$ is not, the number is divisible by $7$ exactly when $7\mid b,$ that is $b=0$ or $b=7.$

For each $a,$ that is $2$ of the $10$ choices of $b,$ a probability of $\dfrac{2}{10}=\dfrac15.$

Thus, the correct answer is E.