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Prove $a + b = c \implies a (\text{mod } N) + b (\text{mod } N) \equiv c (\text{mod } N)$

My attempt:

$a = k_1 \space \text{ (mod n)}$ where $k_1$ is the remainder of $a$.

$b = k_2 \space \text{ (mod n)}$ where $k_2$ is the remainder of $b$.

We have $n \mid a - k_1$ and $n \mid b - k_2$. We can conclude that $$\tag {*} n \mid a - k_1 + b - k_2$$

Reordering $(*)$ gives

$$ n \mid (a+b) - (k_1 + k_2) $$

Since $a + b = c$, we have

$$ n \mid c - (k_1 + k_2) $$

We know that if $n$ divides $x-y$, then $n$ also divides $y-x$, hence

$$ n \mid (k_1 + k_2) - c $$

And therefore

$$a \space (\text{mod } n) + b \space (\text{mod } n) \equiv c \space (\text{mod } n)$$

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Is it correct?

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Yes, your proof is correct. More simply, you can write $a=q_1n+r_1,b=q_2n+r_2$, then $$c=a+b=(q_1+q_2)n+(r_1+r_2)$$ and then the conclusion follows immediately.

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