If X = 11 and Y = 3, then X mod Y = (X + aY) mod Y holds

The correct answer is Option (3) → for all integral values of \(a\)

Adding any multiple of the modulus does not change the remainder. For every integer a, \(aY = 3a\) is a multiple of 3, so \(3a \equiv 0 \pmod{3}\). Hence:

    \((11 + 3a) \bmod 3 = 11 \bmod 3 = 2\)

For example, a = −1, 1, 2 give 8, 14, 17, and each leaves remainder 2 on division by 3. The relation holds for all integers a, not only for even, odd or zero values.

CBSE Applied Mathematics (Class XII):
Congruence Modulo