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