Step 1: Concept Review:
This question focuses on a core modular arithmetic principle: the remainder of `A mod B` is the result of dividing 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: Methodology:
To solve `X mod Y = (X + aY) mod Y`, we will evaluate both sides using the provided values and determine the possible values for 'a' that satisfy the equation.
The fundamental modular arithmetic property is:\[ (A + B) \mod N = ((A \mod N) + (B \mod N)) \mod N \]
Step 3: Detailed Analysis:
Given X = 11 and Y = 3.
Calculate the left-hand side (LHS):\[ \text{LHS} = X \mod Y = 11 \mod 3 \]Dividing 11 by 3 yields a quotient of 3 and a remainder of 2 (11 = 3 * 3 + 2).
Therefore, LHS = 2.
Now, examine the right-hand side (RHS):\[ \text{RHS} = (X + aY) \mod Y = (11 + a \times 3) \mod 3 \]'a' is an integer. The term `3a` is always divisible by 3, irrespective of 'a's value (even, odd, positive, negative, or zero).
Applying the modular arithmetic property:\[ (11 + 3a) \mod 3 = ((11 \mod 3) + (3a \mod 3)) \mod 3 \]We already established that `11 mod 3 = 2`.
Since `3a` is a multiple of 3, its remainder when divided by 3 is always 0.\[ 3a \mod 3 = 0 \]Substitute these results back:\[ \text{RHS} = (2 + 0) \mod 3 = 2 \mod 3 = 2 \]
Step 4: Conclusion:
LHS = 2 and RHS = 2. The equality is satisfied for all integer values of 'a'. Consequently, the statement is universally true for any integer 'a'.