AMC12 2013 A

Answer (E): Let $n$ be a 6-digit palindrome, $m=\frac{n}{11}$, and suppose $m$ is a palindrome as well. First, if $m$ is a 4-digit number, then $n=11m<11\cdot 10^5=10^6+10^5$. Thus the first and last digit of $n$ is $1$. Thus the last digit of $m$ is $1$ and then the first digit of $m$ must be $1$ as well. Then $m\le 1991<2000$ and $n=11m<11\cdot 2000=22000$, which is a contradiction. Therefore $m$ is a 5-digit number $abcba$. If $a+b\le 9$ and $b+c\le 9$, then there are no carries in the sum $n=11m=abcba0+abcba$; thus the digits of $n$ in order are $a$, $a+b$, $b+c$, $b+c$, $a+b$, and $a$. Conversely, if $a+b\ge 10$, then the first digit of $n$ is $a+1$ and the last digit $a$; and if $a+b\le 9$ but $b+c\ge 10$, then the second digit of $n$ is $a+b+1$ if $a+b<9$, or $0$ if $a+b=9$, and the previous to last digit is $a+b$. In any case $n$ is not a palindrome. Therefore $n=11m$ is a palindrome if and only if $a+b\le 9$ and $b+c\le 9$. Thus the number of pairs $(m,n)$ is equal to $$ \sum_{b=0}^{9}\sum_{c=0}^{9-b}\sum_{a=1}^{9-b}1=\sum_{b=0}^{9}(10-b)(9-b). $$ Letting $j=10-b$ gives $$ \sum_{j=1}^{10} j(j-1)=\frac{10\cdot 11\cdot 21}{6}-\frac{10\cdot 11}{2}=330. $$ The total number of 6-digit palindromes $abccba$ is determined by 10 choices for each of $b$ and $c$, and 9 choices for $a$, for a total of $9\cdot 10^2=900$. Thus the required probability is $\frac{330}{900}=\frac{11}{30}$.