Question 877458
<pre>
A A X B B = C C C

(10A+A)(10B+B) = 100C+10C+C
    (11A)(11B) = 111C
         121AB = 111C
           {{{AB/C}}} = {{{111/121}}} 
            

121 and 111 have no common factors other than 1. So
For this to be an equality, the least AB and C could be is 
AB=111 and C=121. No digit C can be 121, since a digit 
is no larger than 9.  So it's not possible

Edwin</pre>