Can the distributivity of the modulo operation be applied to only one operand of an addition? It is known that $(a + b) \bmod n = [(a \bmod n) + (b \bmod n)] \bmod n$, but is it possible that the following are also true?

*

*$(a + b) \bmod n = [(a \bmod n) + b] \bmod n$

*$(a + b) \bmod n = [a + (b \bmod n)] \bmod n$
I've been trying all sorts of tests/combinations (of positive integers, mind you) and it always seems to work. For example, with $a = 7, b = 8, n = 6$:

*

*$(7 + 8) \bmod 6 = 15 \bmod 6 = 3$

*$[(7 \bmod 6) + 8] \bmod 6 = [1 + 8] \bmod 6 = 9 \bmod 6 = 3$

*$[7 + (8 \bmod 6)] \bmod 6 = [7 + 2] \bmod 6 = 9 \bmod 6 = 3$

*$[(7 \bmod 6) + (8 \bmod 6)] \bmod 6 = [1 + 2] \bmod 6 = 3 \bmod 6 = 3$
I've tried it with a dozen different combinations and it always works out. What I would love to know is: am I wrong? If so, is there a good counter-example? If not, how could the veracity of this property be explained/proved, and how come it never shows up in the usual definition of the distributive property of modulo? It's really bugging me, because I'm sure I'm right, but I can't for the life of me find any documentation on this and/or properly demonstrate my case.
Thanks!
 A: As mentioned in comments, this can be shown as in the answer of @BrianM.Scott in Is there a way to simplify this expression $(a + b) \% c$.
However, in your case it seems you take
$$(a + b) \bmod n = [(a \bmod n) + (b \bmod n)] \bmod n\tag{*}$$
for granted, so I will show how to use that. Since $(*)$ holds for all integers $b$, we can substitute $b=c \bmod n$. But then $(*)$ gives us
$$
(a + (c \bmod n)) \bmod n = [(a \bmod n) + ((c \bmod n) \bmod n)] \bmod n.
$$
Since taking remainder by division repeatedly gives the same result, we have $((c \bmod n) \bmod n)=c \bmod n$, and so
$$
(a + (c \bmod n)) \bmod n = [(a \bmod n) + (c \bmod n)] \bmod n.
$$
But right hand side is again by $(*)$ equal to $(a+c) \bmod n$. Thus we have shown
$$
(a + (c \bmod n)) \bmod n = (a+c) \bmod n,
$$
which is one of the equalities you have observed (simply substitute $b$ for $c$). We get also the second equality for free, because $+$ is commutative.
A: Of course it's true.
Suppose $a \mod n = a'$.  That means $0 \le a' < n$ and $a = jn + a'$ for some integer $j$.
Suppose $b \mod n = b'$.  That means $0 \le b' < n$ and $b = kn + b'$ for some integer $k$.
And let's suppose $(a'+b') \mod n = R$ so that $a' +b' = mn + R$.

(Note: because $a', b' < n$ that means either $a' +b' < n$ and $m = 0$ and $R = a'+b'$ or that $n \le a' + b' < 2n$ and $m =1$ and $R = (a'+b') - N$.... but none of that is important.)

So let's look at your expressions:

*

*$(a+b)\mod n=[(a\mod n)+b]\mod n$

$(a+b) = (jn+a')+(kn+b') = (j+k)n + (a'+b') = (j+k)n + mn + R = (j+k+m)n + R$.

So $(a+b)\mod n = R$.

And $(a\mod n) + b = a' + b = a' + kn + b' = kn + (a' + b') = kn + mn + R = (k+m)n + R$.

So $[(a\mod n) + b]\mod n = R$.
So the above is true

*

*$(a+b)\mod n=[a+(b\mod n)]\mod n$
Again $(a+b)\mod n = R$.

And $(a + (b\mod n) = a + b'= jn + a' + b' = jn + mn + R = (j+m)n + R$.

So $[(a+b)\mod n = R$.
