Cause \(\overline{3AA1}⋮9\)

So \(\left(3+2A+1\right)⋮9\Leftrightarrow\left(4+2A\right)⋮9\)

And \(0\le A\le9\) (Cause A is a number which can be represented in a four-digit number).

Make a table and do calculating we have A represent the number 7 which \(\left(4+2\cdot7\right)=18⋮9\)

So the digit that A represents is 7.

Because \(\overline{3AA1}\) divisible by 9

=> 3 + 2A + 1 divisible by 9

But 3 + 1 = 4

=> 2A divide 9 redundancy 5

But 0 \(\le A\le9\)

\(\Rightarrow0\le2A\le18\)

=> 2A = 5 or 2A = 14

If 2A = 5 => A unsatisfactory

If 2A = 14 => A = 7

So A = 7 and \(\overline{3AA1}=3771\)

\(3AA1⋮9\) => \(3+A+A+1⋮9\) => \(4+A.2⋮9\) => \(4+A.2=\left\{9;18;...\right\}\)

If \(A=1\) then \(4+A.2=4+1.2=6\left(wrong\right)\)

...................

We will have \(A=7\)