When the digits of a two-digit number are interchanged, is the resulting number always divisible by 11?

Let the tens digit be $a$ and the units digit be $b$.
The original number can be written as $10a + b$.
Here, $a$ is an integer from 1 to 9, and $b$ is an integer from 0 to 9.

When the digits are interchanged, the new tens digit is $b$ and the new units digit is $a$.
The new number can be written as $10b + a$.

The sum of the original number and the new number is:

Combine like terms:

Factor out 11 from the sum:

Since the sum is $11(a + b)$, it is always a multiple of 11. Therefore, the resulting number will always be divisible by 11.

Let's check the given options:

(a) 9

(b) 3

(c) 11

(d) 5

The sum $11(a+b)$ is always divisible by 11. It is not necessarily divisible by 9, 3, or 5 for all possible values of $a$ and $b$. For example, if the original number is 12, then $a=1, b=2$.
Original number = 12
New number = 21
Sum = $12 + 21 = 33$.
$33 = 11(1+2) = 11(3)$.
33 is divisible by 11 and 3, but not by 9 or 5.
If the original number is 23, then $a=2, b=3$.
Original number = 23
New number = 32
Sum = $23 + 32 = 55$.
$55 = 11(2+3) = 11(5)$.
55 is divisible by 11 and 5, but not by 9 or 3.
The only number that consistently divides the sum is 11.

The correct option is (c).

The final answer is $\boxed{\text{11}}$.