Question

A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base $10$ with no leading zeros. A $6$ -digit palindrome $n$ is chosen uniformly at random. What is the probability that $\dfrac{n}{11}$ is also a palindrome?

$\dfrac{8}{25}$

$\dfrac{33}{100}$

$\dfrac{7}{20}$

$\dfrac{9}{25}$

$\dfrac{11}{30}$

Solution:

Let $m = \dfrac{n}{11}.$ A $4$ -digit $m$ leads to a contradiction, so $m$ is a $5$ -digit palindrome $\overline{abcba}.$

Writing $n = 11m,$ there are no carries exactly when $a + b \le 9$ and $b + c \le 9,$ and only then is $n$ a palindrome. 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.

2013 AMC 12A Problem 22

Solution:

Problem 22 in Other Years