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

Given $X = 11$ and $Y = 3$.

The modulo operation $X \mod Y$ gives the remainder when dividing $X$ by $Y$.

$11 \div 3 = 3$ with remainder $2$

So $11 \mod 3 = 2$

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Testing $(11 + a \times 3) \mod 3$ for different values of $a$:

For $a = 1$:

$(11 + 1 \times 3) \mod 3$

$= 14 \mod 3$

$= 2$

For $a = 2$:

$(11 + 2 \times 3) \mod 3$

$= 17 \mod 3$

$= 2$

For $a = 0$:

$(11 + 0 \times 3) \mod 3$

$= 11 \mod 3$

$= 2$

For $a = -1$:

$(11 + (-1) \times 3) \mod 3$

$= 8 \mod 3$

$= 2$

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Adding any multiple of $Y$ to $X$ does not change the remainder when dividing by $Y$.

Since $aY$ is a multiple of $Y$:

$aY \mod Y = 0$

Therefore:

$(X + aY) \mod Y$

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

$= (X \mod Y + 0) \mod Y$

$= X \mod Y$

This holds for all integers $a$ (positive, negative, or zero).

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The equation $X \mod Y = (X + aY) \mod Y$ holds for all integral values of $a$.