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

for all integral values of $a$

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

Given: $X=11$, $Y=3$.

Check the identity:

$X \bmod Y = (X + aY)\bmod Y$

Compute:

$11 \bmod 3 = 2$

Now consider $11 + a\cdot 3 = 11 + 3a$.

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

$3a \equiv 0 \pmod{3}$ for every integer $a$

So:

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

The identity holds for **every integer value of** $a$.

Final answer: for all integral values of $a$