Wow! Crazy cryptogram!

Since the numbers begin with $A$ and $B$ , neither of them is zero.

The multiplication of the units $B\cdot A = ?B$ gives three possibilities:

• $A = 1$

• $A = 6$ and $B$ is even;

• $B = 5$ and $A$ is odd.

The first case is easily ruled out because the product would have no more than three digits.

The second case fails, too. Since $6\times 6=36$ , we would need $B$ to be equal to 4; but $64\times 66 = 4224$ is not quite right.

The third case works. In order for the product to start in 5, we need $A\times A\approx 50$ , which implies $A = 7$ . It is easy to see that this works: $\boxed{75}\times 77 = 5775$ .

(10A+B)(10A+A)= (10A+B)(11A)
110A^2+11AB=1000B+100A+10A+B
110A^2 +11AB= 1001B+110A
110A(A-1)= B(1001-11A)

1)For A=1, B must be zero.
This condition fulfills the given problem but we do not write zero at thousandth place when it is a three digit number.

So to satisfy the given condition and

110A(A-1)=B(1001-11A)
B must be multiple of five<10 (because 1001-11A is not a multiple of 10 unless A=1)
ie B=5
So A=7

Looking at the units digits, we need the units digit of $A\cdot B$ to be $B.$

Then, it's straightforward to check the possibilities going through the values of $B.$

We find $\overline{AB} = 75$ gives $75 \cdot 77 = 5775.$

$\Rightarrow \overline{AA}×\overline{AB}=\overline{BAAB} \\ \Rightarrow 110A^2+11AB=1001B+110A \\ \Rightarrow 10A^2+AB=91B+10A \\ \Rightarrow 10A^2-10A=91B-AB \\ \Rightarrow 10A(A-1)=B(91-A) \\ \Rightarrow B=\frac{10A(A-1)}{(91-1)} \\ \text{The only values of A and B which will yield a 4 digit number are 7 and 5}. \\ \Rightarrow \overline{AB}=\boxed{75}$

A has to be higher than 2 as 33*31 is the lowest 4 digit product.

Then B cannot be 1 | B*A is not B for B=1

Thus we try A= 4 and greater , A=5 is not true , so A = 4,6,7,8,9

A=4 does not have B true for any digit for the required units place of A

A=6 has a couple of candidates but they do not land a corresponding value in the thousands (B=2,4,8), 77*75 = 5775

The answers for A=6 for 62 64 and 68 are all interesting as well.

By long division we have $\frac{\overline{BAAB}}{11}$ = $100B+10(A-B)+B$

What we are multiplying from the cryptogram: $\overline{AB} \times \overline{AA} = (10A+B) \times 11A = 11(10A^2+AB)$

Rearranging we have

$100B+10(A-B)+B = (10A^2+AB)$

$B(91-A) = 10A(A-1)$

$B = \frac{10A(A-1)}{91-A}$

Since $A$ and $B$ are single digit integers, try each one to see only $A = 7$ will give $B$ as integer.

So $B = \frac{10(7)(6)}{91-7}=\frac{420}{84}=5$

so $\overline{AB}=75$

The question is ambiguous. It should simply say: 'Write down the two-digit number AB.'

Not a solution, just an issue with the wording. I find the use of the word "product" confusing because, to me, the product is clearly BAAB. I would have it ask, "What is the two-digit number in the multiplier/multiplicand AB above?" or simply, "What is the two-digit number AB?"

00 is also a solution