Numbers... without numbers 3

C*C=AB so 9<C^2<100. So just find the squares of 4,5,6,7,8,9 and one of these is the solution.

It is given that the number $\overline{AB}$ is two digits and a perfect square.

There are only six numbers that are two-digit perfect squares: $16, 25, 36, 49, 64, 81$ .

From the problem statement, $\overline{AB}$ must also satisfy the condition: $(A + B)^2 = \overline{AB}$ .

Testing each of the six perfect squares, only $81$ satisfies this condition: $(8 + 1)^2 = 81$ .

Hence, $\overline{AB} = 81$ .

Because AB is a 2-digit number:
$9<\overline { AB } ={ C }^{ 2 }={ (A+B) }^{ 2 }<100$
So AB must be 16, 25, 36, 49, 64 or 81 (squares of 4,5,6,7,8,9). And we have $3<A+B<10$
So we get rid of 49 and 64, as their A+B > 10.
${ (A+B) }^{ 2 }={ (8+1) }^{ 2 }={ 9 }^{ 2 }=81$
So it's 81.
(I know it looks like Alvi Newaz's solution, I just want to make it clearer and somehow easier)