Let two-digit of A = x => 10x
Let two-digit of B = y
A = 10x + y
(10x + y) = 2y + B
10x + y - 2y = B
10x - y = B
$$Therefore,\ product\ AB=\left(10x+y\right)\left(10x-y\right)=100x^2-y^2$$
Question: What is the hundreds digit of the product of A and B?
Statement (1): The tens digit of A is prime.
i.e x is prime
So, if x=2, and y=3
$$Then\ AB=100\left(2^2\right)-3^2=400-9=391$$
The hundreds digit of AB = 8
The information provided cannot give definite hundred of digits for the product AB, hence, statement 1 is NOT SUFFICIENT.
Statement (2): Ten is not divisible by the tens digit of A.
i.e x could be either 3, 4, 6, 7, 8, or 9 because they cannot divide 10.
If x=3 and y=1,
$$then\ AB=100\left(3^2\right)-1^2=900-1=899$$
The hundreds digit of AB = 8
If x=6 and y=1,
$$then\ AB=100\left(6^2\right)-1^2=3600-1=3599$$
The hundreds digit of AB = 5
The possible values of AB yield different hundreds digit for product AB, hence, statement 2 is NOT SUFFICIENT.
Combining both statements together;
x is a prime number that can divide 10, so x=3 or x=7
If x=3 and y=1,
$$then\ AB=100\left(3^2\right)-1^2=900-1=899$$
If x=7 and y=2,
$$then\ AB=100\left(7^2\right)-2^2=4900-4=4896$$
In both cases, the hundred digit of product AB=8 irrespective of the valye of y.
Hence, both statements combined ARE SUFFICIENT. Thus, the correct answer is option C.
