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A A X B B = C C C
(10A+A)(10B+B) = 100C+10C+C
(11A)(11B) = 111C
121AB = 111C
=
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