AA + BB + CC = BAC

Can someone help - need numbers substituting the letters above.

I am unable to figure out. Thank you.

11(A+B+C) =100*B+10*A+C
A=89B-10C
you need to find the integers from 0 to 9 forA, B, C to solve above eqn.
and I know none of them?????
what about A=0 B=0 C=0(that's obvious)

__________________

Dont dance here !

I was under the impression that AA had to be the same numbers, i.e. 11 or 22

and likewise with BB and CC

.. am I wrong in thinking this way?

No you are not
you must have noticed
00,11,22,33,44...99
are all multiples of 11
so
I took 11 common from
AA+BB+CC
which gave
(A+B+C) x 11 = ....
and i tried it further
since there was no restriction given on
A, B,C
So I took 0 for each one of them
and it proved the result that's on my earlier post

__________________

Dont dance here !

I cant find any oter soln trivial way.....(I am editing this post later on)
anyway I am Ada"""r"""sh

__________________

Dont dance here !

the answer is

99
11
88

which equals 198.