This is an exercise from a textbook in Portuguese.
Choose a symbol (number) $abc$ in the decimal system ($a$ being the hundreds digit, $b$ the tens digit, and $c$ the units digit), in a way that the hundreds digit $a$ and the units digit $c$ differ by at least $2$ units. Consider the numbers $abc$ and $cba$ and subtract the smaller from the bigger in order to get a number $xyz$. Show that the sum of $xyz$ with $zyx$ is 1089.
I know that $abc=a10^{2}+b10+c$, $cba=c10^{2}+b10+a$, and $c=a+i$, $i\geq 2$ (or $a=c+i$, $i\geq 2$). So if $c=a+i$, $i\geq 2$, then $cba=(a+i)10^{2}+b10+(c-i)=a10^{2}+b10+c+(i10^{2}-i)=a10^{2}+b10+c+(99i)\;.$ So $xyz=cba-abc=99i$, or $9 \mid xyz$ wich give me that $x+y+z=9k$, and $11 \mid xyz$ wich give me a condition, but I still can use this.
I would appreciate your help! Sorry for the lousy title.