Question 943984
The answer is not {{{18}}} , because {{{a+b<18}}} .
Since {{{a}}} and {{{b}}} are digits,
{{{system {{{a,<=9,b<=9}}}--->{{{a+b<=18}}} , 
but {{{a+b=18}}} means {{{a=b=0}}} ,
and then {{{a}}} and {{{b}}} would not be "distinct digits".
The value of the two-digit number ab is {{{10a+b}}} .
The value of the two-digit number ba is {{{10b+a}}} .
The difference of ab and ba is
{{{abs(10a+b-(10b+a))=abs(10a+b+10b-a)=abs(9a-9b)=abs(9(a-b))=9*abs(a-b)}}} .
Since {{{9*abs(a-b)}}} is the highest common factor (or greatest common factor) of ab and ba,
we know that {{{9}}} is a factor of ab and ba.
As a consequence, {{{9}}} is a factor of {{{a+b}}} .
In other words, {{{a+b}}} is a multiple of {{{9}}} .
The only {{{a+b}}} multiple of {{{9}}} such that {{{a+b<18}}} is {{{highlight(a+b=9)}}} .