Intermediate Mathematics-4.

Let $n=100a+10b+a = \overline{aba}$ and $m = n+192=\overline{cddc}$ , since they are palindromes. Therefore,

\[\begin{array} {} & & a & b & a \\ + & _{\color{red}1} & 1 & 9 & 2 \\ \hline & c & d & d & c \end{array} \]

where the small red $\color{#D61F06}1$ indicates carry-forward of $a+1$ .

For $m$ to be a 4-digit integer, $a$ can either be 8 or 9. We also note that the thousand digit of $m$ , which is $c$ , can only be 1. Therefore, the unit digit of $m$ must also be $c=1$ . This means that $a=9$ , so that $a+2 = 11$ . Then we have:

\[\begin{array} {} & & 9 & b & 9 \\ + & _{\color{red}1} & 1_{\color{red}1} & 9_{\color{red}1} & 2 \\ \hline & 1 & \color{blue}1 & \color{blue} b & 1 \end{array} \]

Since $m$ is a palindrome, $\color{#3D99F6}b=1$ , $\implies n = 919$ and $919+192 = 1111$ . Therefore, the sum of digits of $n$ is $9+1+9 = \boxed{19}$ .

$\overline{aba}+192=\overline{xyyx}$

$a$ can only be $8$ or $9$ , otherwise $\overline{aba}+192$ would be a three digit number.

$\overline{aba}+192=\overline{xyyx} \implies 101a+10b+192=1001x+110y$

so

$10|101a+192-1001x \implies 10|a+2-x$

if $a=8$ then $x=0$ , which is not an option. So, $a=9$ and $x=1$

the rest would be solving a rather easy Diophantine equation.

$110y-10b=100$

so, $y=1$ and $b=1$ .

$919+192=1111$

$9+1+9=19$

Note that $808+192=1000$ , and that $808+x$ is also a palindrome, if $x$ is in the set $\{10,20,30,40,50,60,70,80,90,101,111,121,131,141,151,161,171,181,191\}$ . Only $1000+111$ is a palindrome, because you need a $1$ at the end of the number to match the $1$ at the beginning of $1000$ , and you need the first two digits in that number to equal each other, which is only satisfied by $111$ . Therefore, $n=808+111=919$ , and $9+1+9=19$ , as shown above.