Step 1: Understanding the Concept:
This question assesses a core principle of modular arithmetic. The operation `A mod B` yields the remainder of the division of A by B. The relevant property is `(A + kB) mod B = A mod B`, where k is any integer. This holds because `kB` is a multiple of `B`, and adding a multiple of the divisor does not affect the remainder.
Step 2: Key Formula or Approach:
We must evaluate both sides of the equation `X mod Y = (X + aY) mod Y` using the provided values and determine the values of 'a' for which the equality is valid.
The fundamental property of modular arithmetic is:
\[ (A + B) \mod N = ((A \mod N) + (B \mod N)) \mod N \]
Step 3: Detailed Explanation:
The given values are X = 11 and Y = 3.
First, let's compute the left-hand side (LHS) of the equation:
\[ \text{LHS} = X \mod Y = 11 \mod 3 \]
Dividing 11 by 3 results in a quotient of 3 and a remainder of 2 (since \(11 = 3 \times 3 + 2\)).
Therefore, LHS = 2.
Now, let's analyze the right-hand side (RHS) of the equation:
\[ \text{RHS} = (X + aY) \mod Y = (11 + a \times 3) \mod 3 \]
Here, 'a' represents any integer. The term `3a` is invariably a multiple of 3, irrespective of whether 'a' is even, odd, positive, negative, or zero.
Applying the property of modular arithmetic:
\[ (11 + 3a) \mod 3 = ((11 \mod 3) + (3a \mod 3)) \mod 3 \]
We already established that `11 mod 3 = 2`.
Since `3a` is a multiple of 3, its remainder when divided by 3 is always 0.
\[ 3a \mod 3 = 0 \]
Substituting these values back into the RHS expression:
\[ \text{RHS} = (2 + 0) \mod 3 = 2 \mod 3 = 2 \]
Step 4: Final Answer:
We observe that LHS = 2 and RHS = 2. The equality holds true for any integer value of 'a'. Consequently, the statement is valid for all integral values of 'a'.