A palindrome is a number that is the same when read forwards and backwards, such as $43234$. Find the number of $4$-digit palindromes that are divisible by $13.$

Answer 1

Let the palindrome in question be $abba$, where $a$ and $b$ are digits. By Rule 1 of Divisibility by $13$, the palindrome is divisible by 13 if and only if $10a+b \equiv 10b+a \pmod{13},$ which gives us $9(a-b) \equiv 0 \pmod{13}.$

Since $9$ and $13$ are relatively prime, it follows that $a - b$ is divisible by 13.

Since $a$ and $b$ are digits, $a-b$ can only equal $-12,$ $-11$, $\dots$, $-1$, $0$, $1$, $\dots$, or $11.$

Thus, there are $11 - (-12) + 1 = \boxed{24}$ palindromes divisible by 13.