2013 AMC 12A Problem 22

Show $m = \dfrac{n}{11}$ must be a $5$-digit palindrome $\overline{abcba}$

The product $11m$ is a palindrome exactly when $a + b \le 9$ and $b + c \le 9;$ count those

Let $m= \frac{n}{11}.$ If $m$ had four digits, then $n<110000,$ so the first and last digits of the six-digit palindrome $n$ would both be $1.$ This forces the first and last digits of the palindromic $m$ to be $1,$ hence $m<2000$ and $n<22000,$ a contradiction. Therefore $m$ is a five-digit palindrome $\overline{abcba}.$

Writing $n=11m=\overline{abcba0}+\overline{abcba},$ no carries occur exactly when $a+b\le9$ and $b+c\le9;$ the resulting digits are $a,a+b,b+c,b+c,a+b,a.$ If $a+b\ge10,$ the leading and trailing digits differ; if only $b+c\ge10,$ the next pair differs. Thus the conditions are also necessary. The number of valid $m$ is $$\sum_{b=0}^{9}(10 - b)(9 - b) = 330.$$

There are $9\cdot 10^2 = 900$ six-digit palindromes, so the probability is $\dfrac{330}{900} = \dfrac{11}{30}.$

Thus, the correct answer is E.