Answer: Option B. -> 5
:
B
The sum of three A's is such anumber that hasA in its ones place.
Therefore, the sum of two A's must be a number whose ones digit is 0.
This happens only for A = 0 and A = 5
If A = 0, then the sum is 0 + 0 + 0 = 0, which makes B = 0.
Given that B is a natural number, thereforeB cannot be equal to 0. So we don't considerthis possibility.
Hence, A = 5$$\begin{array}{c}5\\ +5\\ +5\\ 15\end{array}$$
Therefore, A= 5 and B= 1

Answer: Option C. -> 11
:
C
Let a be the tens place digit and b be the units placedigit, then the original number is 10 x a + 1 x b.The new number after reversing the digits is 10 x b + 1 x b.Therefore the sum will be:
(10a + b) + (10b + a) = 11a + 11b
= 11 (a + b)
Which is divisible by 11.

Answer: Option B. -> No
:
B
Let's take a two digit number 'ab'. Its general form is10×a + 1×b. Whenwe reverse the digits,the new number will be of the form 10×b + 1×a.
If we add boththe numbers we get
(10×a + 1×b) + (10×b + 1×a)
=10a + b + 10b + a
= 11(a+b)
Hence when divided by 11 it will notleave any remainder.

Answer: Option C. -> 5
:
C
Ones digit of 3 $\times$ A is A. So, it must be either A = 0 or A = 5.
Now, if B = 1, then BA $\times$B3 would at most be equal to 15$\times$ 13; (Taking A= 5 to find the maximum value of the product)that is,it would at most be equal to 195. But the product here is 57A, which is more than 500. So we cannot have B =1.
If B = 3, then BA $\times$ B3 would be at least30$\times$ 33; (Taking A = 0 to find the minimum value of the product) that is990. But 57A is less than 990. So, B cannot be equal to 3.
Putting these two facts together, we get B = 2.
So, the multiplication is either 20 $\times$ 23, or
25 $\times$ 23.
The first possibility fails, since
20 $\times$ 23 = 460. But, the second one works out correctly, since
25 $\times$ 23 = 575. So the answer is A = 5 andB = 2.
$$\begin{array}{c}25\\ \times 23\\ 575\end{array}$$

Answer: Option D. -> 87
:
D

On dividing 487 by 100, the remainder is the number formed by the last 2 digits of the given number.  Hence it is 87 in this case.

Answer: Option C. -> 5
:
C

The sum of the digits of 152875 is 28 which is not a multiple of 3, 152875 is not divisible by 3.
28 is not a multiple of 9 so the number is also not divisible by 9.
Number formed by last two digits is 75 and it is not divisible by 4. Therefore, the number is not divisible by 4.
The last digit of given number is 5, hence 152875 is divisible by 5.

Answer: Option C. -> 6
:
A, B, and C

The sum of the digits of the given number is 4 + 5 + 6 + 3 + 1 + 8 = 27 which is divisible by 3.
Hence, the given number is divisible by 3.
The last digit of the number is even. Hence, it is divisible by 2.
Since the number is divisible by both 3 and 2, it is divisible by 6.
A number is divisible by 11 , if  the difference between the sum of its digits in even places and the sum of its digits in odd places is either 0 or divisible by 11.
Here, sum of digits in even places
= 1 + 6 + 4 = 11
Sum of digits in odd places
= 8 + 3 + 5 = 16
Difference = 16 -11 = 5 which is not divisible by 11. 
$\therefore$ The given number is not divisible by 11.

Answer: Option A. -> True
:
A

Let abc be the number.
Thus abc = 100a + 10b + c
The number obtained by reversing the digits is cba.
cba = 100c + 10b + a
Adding the two numbers we get, abc + cba = (100a + 10b + c) + (100c + 10b + a) = 101a + 20b + 101c
Since middle digit is 0, the sum will be 101(a+c), which is divisible by 101.

Answer: Option A. -> 3
:
A and D

Let the number be ab.
Here we have 'a' in tens place and 'b' in units place. So we can write the number as 10a + b. If we reverse it we get 10b+a.

(10a + b) – (10b + a) = 10a + b – 10b – a

                                  = 9a – 9b = 9(a – b).

Hence the number is divisible by 9.
Additionally the resulting number will also be divisible by factors of 9 which is 1 and 3 in this case. 
Hence the number obtained will be divisble by 3 and 9.

Answer: Option A. -> $({10}^2\times 9)+({10}^1\times 0)+({10}^0\times 8)$
:
A

908 can be written as
$(100\times 9)+(10\times 0)+(1\times 8)$
= $({10}^2\times 9)+({10}^1\times 0)+({10}^0\times 8)$