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**Answer: A, C, D**

Given $a \equiv b \text{ (mod n)}$ i.e. $(a-b)\% n = 0$

__Option A: True__

$a \equiv b-3n \text{ (mod n)} = (a-b+3n)\%n = ((a-b)\%n +(3n)\%n)\%n = (0+0)\%n = 0$

__Option B: False__

$a \equiv b+k \text{ (mod n)} = ((a-b)\%n - (k\%n))\%n = (-k)\%n$

Now since $k$ can be any integer, $k$ might not be a multiple of $n$.

__Option C: True__

$a + k \equiv b+k \text{ (mod n)} = (a+k-b-k)\%n = (a-b)\%n = 0$

__Option D: True__

$a+5n \equiv b-3n \text{ (mod n)} = (a+5n-b+3n)\%n = ((a-b)\%n + (8n)\%n)\%n = (0+0)\%n = 0$