Which of the following are correct?

(A) If a ≡ b(mod n), then -a ≡ -b (mod n)

(B) If a + b = c, then a(mod n) + b(mod n) ≡ (a + b + c)(mod n).

(C) If a ≡ b (mod n), then ka ≡ kb(mod n), ∀ k ∈ I.

(D) If a ≡ b (mod n), then (a + k) ≡ (b + k)(mod n), ∀ k ∈ I.

Choose the correct answer from the options given below:

When $a \equiv b \pmod{n}$, it means $n$ divides $(a - b)$ completely.

---

For statement (A): If $a \equiv b \pmod{n}$, then $-a \equiv -b \pmod{n}$

If $a \equiv b \pmod{n}$, then $n$ divides $(a - b)$.

$(-a) - (-b)$

$= -a + b$

$= -(a - b)$

Since $n$ divides $(a - b)$, it also divides $-(a - b)$.

Therefore, $-a \equiv -b \pmod{n}$.

Statement (A) is correct.

---

For statement (B): If $a + b = c$, then $a \pmod{n} + b \pmod{n} \equiv (a + b + c) \pmod{n}$

Testing with $a = 2$, $b = 3$, then $c = 5$ and $n = 3$:

Left side: $2 \pmod{3} + 3 \pmod{3}$

$= 2 + 0$

$= 2 \pmod{3}$

Right side: $(2 + 3 + 5) \pmod{3}$

$= 10 \pmod{3}$

$\equiv 1 \pmod{3}$

Since $2 \not\equiv 1 \pmod{3}$, the statement is false.

Statement (B) is incorrect.

---

For statement (C): If $a \equiv b \pmod{n}$, then $ka \equiv kb \pmod{n}$, $\forall k \in \mathbb{Z}$

If $a \equiv b \pmod{n}$, then $n$ divides $(a - b)$.

$ka - kb$

$= k(a - b)$

Since $n$ divides $(a - b)$, it also divides $k(a - b)$ for any integer $k$.

Therefore, $ka \equiv kb \pmod{n}$.

Statement (C) is correct.

---

For statement (D): If $a \equiv b \pmod{n}$, then $(a + k) \equiv (b + k) \pmod{n}$, $\forall k \in \mathbb{Z}$

If $a \equiv b \pmod{n}$, then $n$ divides $(a - b)$.

$(a + k) - (b + k)$

$= a + k - b - k$

$= a - b$

Since $n$ divides $(a - b)$, it divides $(a + k) - (b + k)$.

Therefore, $(a + k) \equiv (b + k) \pmod{n}$.

Statement (D) is correct.

---

Statements (A), (C), and (D) are correct.