2010 AMC 10B Problem 21

A four-digit palindrome reads the same both ways, like $3443$. It has the form $abba$: first digit $a$ from $1$ to $9$ and middle digit $b$ from $0$ to $9$. That makes $9 \cdot 10 = 90$ palindromes.

1. Split by place value. Take $3443$:
$$3443 = 3000 + 400 + 40 + 3 = 3 \cdot 1001 + 4 \cdot 110.$$
In the same way every palindrome $abba$ equals $1001a + 110b$.
2. The first part is always a multiple of $7$, because $1001 = 7 \cdot 143$ (in fact $1001 = 7 \cdot 11 \cdot 13$).
3. So the palindrome is a multiple of $7$ exactly when $110b$ is. Since $110 = 2 \cdot 5 \cdot 11$ has no factor of $7$, the factor $7$ must come from $b$ itself. So $b$ is a multiple of $7$: $b = 0$ or $b = 7$.
4. For every first digit $a$, $2$ of the $10$ middle digits work. The probability is $\frac{2}{10} = \frac15$.

Check two examples: $1771 = 7 \cdot 253$ and $3003 = 7 \cdot 429$.

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

Writing a number in terms of its digits (place value) shows which digits actually matter for a divisibility question. The fact $1001 = 7 \cdot 11 \cdot 13$ is worth memorizing: it is why every four-digit palindrome is a multiple of $11$, and why here only the middle digit decides divisibility by $7$.

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