SOLUTION: Find A, B and C in the multiplication table. A B C x A A = A C 6 C

A B C x A A =  A C 6 C

A must be 1 because if it were as much as 2, 
the smallest ABC could be would be 201. Then
we'd have
201 x 22 = 4422 but the product would then have to 
start with A=2.

So A=1

So if we perform the multiplication we'd have this:

          1BC
          x11
          1BC
         1BC0
         1C6C

From the 2nd from the right column in the 
addition part, either B+C=6 or B+C=16

if B+C=6 then the 2nd column from the left would be 1+B=C
So we'd have the system of equations 
 But that has solution B=5/2, C=7/2,
which aren't digits.

So B+C=16  and the only digits that have sum 16 are 7 and 9
Then there'd be 1 to carry to the 2nd column from the left,
so 1+1+B=C or 2+B=C, so B=7 and C=9.

So the solution is:

          179
          x11
          179
         1790
         1969

Edwin