The last digit satisfies $A \times A = \overline{?A},$ which is only the case if $A = 1, 5, 6$ .

Immediately we rule out $A = 1$ (because then the product would have only 3 digits). That leaves $5 \times 555 = 2775,$ which is too small; and $6 \times 666 = 3996,$ which is the solution. Thus, $B = \boxed 9$ .

$\begin{aligned}\text{We have,}\\ \large\overline{\color{#D61F06}aaa}\times {\color{#D61F06}a}&=\overline{3bb\color{#D61F06}a}\\ \large\implies \color{#D61F06}a^{2}\color{#333333}\times111&=\overline{3bb\color{#D61F06}a}\hspace{4mm}\color{#3D99F6}\small(1)\\ 3000&\leq\overline{3bb\color{#D61F06}a}\leq3999\hspace{4mm}\color{#3D99F6}\small(2)\\ \text{The only possible }&\text{value of }\color{#D61F06}a \color{#333333}\text{ satisifying } \color{#3D99F6} (1) \color{#333333}\text{ and }\color{#3D99F6} (2) \text{ is } \color{#D61F06}a\color{#333333}=6 \\ \large\implies \color{#D61F06}a^{2}\color{#333333}\times111&=36\times111=3996\\ \implies \color{#EC7300}&\boxed{\color{#333333}b=9}\end{aligned}$

(100a+10a+a)a=3000+100b+10b+a;

111a²-a=3000+110b;

b=(111a²-a-3000)/110;

111a²-a>3000;

minimum a=6 then b=9;

We see that the upper product displays $a^2=\overline{3a}$ so it same as $aa=30+a \implies a(a-1)=30$ which only 6 solve the solution

Hence, $\color{#D61F06}\boxed{\large{a=6}}$

The upper product displays $\overline{aaa} \times a=\overline{3bba}$ so it same as $3000<111a \times a<3999 \implies 3000<111a^2<3999$ .Because it contains 111 which is part of $10^n+10^{n-1}+10^{n-2}+\dots+10^3+10^2+10+1$ so that means we have to take 3 from $\overline{3a}$ then plus $a$ which is 9

Hence, $\color{#3D99F6}\boxed{\large{b=9}}$

At first, we have to know which digit A represents. We know A x A = A. So we´re searching a one digit number, which has itself as the last digit if we square it.

There are only four numbers, which fulfill the condition: 1, 5, 6 and 0. The second condition is that the leading digit must be 2 or 3 because of the leading digit of the calculations' solution is, not the B and it's a 3 Looking back at the remaining numbers and at the table shows that only one number remains: 6. Finally, the solution of the problem is the sum of both digits of 6 squared. Looking back at the table again reveals 6 squared is 36 and 3 + 6 = 9

Out of 1, 5 and 6 as the possible value for A, 6 satisfies the requirements and 666x6=3996 gives 9 as the value for B, Therefore, Answer = 9

I didn't want to solve this problem using logic, so I just made a Python 3.7 program and solved it:

x = 0;
for x in range(0, 9):
    num = (x*100) + (x*10) + x;  
    print(str(num) + " * " + str(x) + " = " + str(int(num) * int(x))

Using the program, it says that '666 * 6 = 3996', therefore:

B = 9

666*6= 3996 ... A=6 B=9

A must be a number when multiplied by itself, should end up being a number with the last digit being the same number.

That means A can equal 1 or 6 because...

1 x 1 = 1 6 x 6 = 36

If you substitute A with both 1 and 6, 6 ends up being the number that gives the answer with a 3 as the first digit. After multiplying the answer is 3996, meaning B equals 9.

Could you not have just thought of it as the choices having to be a perfect square, and that it 3 has to be a factor of one of the choices?

The last digit of “3 B B A” (A) = the last digit of A x A which we know simply from the basic method of long multiplication. This is only true for 3 numbers: 1, 5 and 6. We know 111 x 1 = 111, hence one is impossible as the product begins with 3. 555 x 5 starts with a 5, which we know from the method of long multiplication. 666 x 6 does begin with 3 as 6 x 6 = 36 and long multiplication states that we start with that 3 from 36. This is pretty cool as we see how 555 x 5 and 666 x 6 start with the first letter of their squares and end with the last letter of their square. It’s also true that the numbers in between the start and end = the sum of the numbers of the square of the single digit eg: 6 x 6 = 36, so the other digits (B) = 3 + 6= 9 therefor B = 9 ( its also true the total amount of digits = the amount of numbers in the sum) Let’s prove the above statement for understanding.

Take 4

4 x 4 = 16 Therefor we know 444 x 4 = 1776, no calculator needed. Let’s look at long multiplication, the most simple math technique taught to grade 1s and 2s 444

x4

16+ 160 + 1600 since we add a zero each time when we multiply each miner I the top by the number in the bottom. From here it’s easy to see how the Coloums will add up to prove our statement.

666 × 6 = 3996, so A = 6 and B = 9

If the number would have been 111, 222 or 333 then multiplying it by 1,2 and 3 respectively will give us a 3 digit number. So, they can not be the numbers. Multiplying 444 and 555 with 4 & 5 respectively will give the numbers which doesn't start's with 3. But 666×6 = 3996 which matches exactly with the given form. Hence, by comparing we get A=6 & B=9.

To work out B we first need to work out A. So looking at the equation, A x A must end in A. The only whole numbers A can be is "6" or "5" because 6 x 6 = 36, and
5 x 5 = 25. It can't be 5 because 555 x 5 = 2775, and we need the number in the thousands place value to be "3" which it isn't. So lets try "6" 666 x 6 = 3996. So B = 9