2010 AMC 10B Problem 21

A four-digit palindrome has the form $\overline{abba}$ with digits $a \in \{1,\dots,9\}$ and $b \in \{0,\dots,9\}$, so there are $90$ of them. In terms of place value,
$$\overline{abba} = 1000a + 100b + 10b + a = 1001a + 110b.$$

Since $1001 = 7 \cdot 11 \cdot 13$, the term $1001a$ is always a multiple of $7$. So the palindrome is divisible by $7$ exactly when $110b$ is. Because $\gcd(110, 7) = 1$, this means $7 \mid b$, i.e. $b = 0$ or $b = 7$.

For each of the $9$ choices of $a$, exactly $2$ of the $10$ choices of $b$ work, so the probability is $\frac{2}{10} = \frac15$.

The answer is $\boxed{\textbf{(E)}\ \dfrac{1}{5}}$.

Writing a digit pattern as a linear combination of its digits reveals which digits actually matter for a divisibility question. The factorization $1001 = 7\cdot 11\cdot 13$ is worth memorizing; it is why four-digit palindromes are always divisible by $11$ and why their divisibility by $7$ depends on the middle digit alone.

• Trying to list all 90 palindromes and test each, or forgetting that b = 0 is allowed (giving 1/10 instead of 1/5).
• Guessing $\frac17$ on the grounds that "one in seven numbers is divisible by $7$," which ignores the structure of palindromes.