Which of the following are correct?

(A) If $\text{a ≡ b(mod n)}$, then $\text{-a ≡ -b (mod n)}$
(B) If $\text{a + b = c}$, then $\text{a (mod n) + b(mod n) ≡ (a+b+c)(mod n)}$.
(C) If $\text{a ≡ b (mod n)}$, then $\text{ka ≡ kb(mod n),∀ k∈ I}$.
(D) If $\text{a ≡ b (mod n)}$, then $\text{(a + k) ≡ (b + k) (mod n), ∀ k∈l}$.

Choose the correct answer from the options given below:

(A), (C) and (D) only

The correct answer is Option (3) → (A), (C) and (D) only

(A) True. If $a\equiv b\pmod{n}$ then $n\mid(a-b)$, so $n\mid-(a-b)=(b-a)$ and $-a\equiv -b\pmod{n}$.

(B) False. If $a+b=c$ then $(a+b+c)=2c$, and $a\pmod{n}+b\pmod{n}$ need not be congruent to $2c\pmod{n}$ in general.

(C) True. If $a\equiv b\pmod{n}$ then $n\mid(a-b)$, so $n\mid k(a-b)$ and $ka\equiv kb\pmod{n}$ for all integers $k$.

(D) True. If $a\equiv b\pmod{n}$ then $n\mid(a-b)$, hence $n\mid\big((a+k)-(b+k)\big)$ and $(a+k)\equiv(b+k)\pmod{n}$ for all integers $k$.