Properties of modular arithmetic clarification The sum of cubes of any three consecutive integers is divisible by $9$. i.e $(n^3 + (n+1)^3 + (n+2)^3)\mod9=0$ for every $n\in\mathbb{Z}$. 
They say it is sufficient to check out for values of $n=0,1,2,3,...8$ due to properties of modular arithmetic. Why this is so ?
Thanks
 A: This happens because of the following properties, which you can easily verify:
$$ (a + b) \mod c \equiv ((a \mod c) + (b \mod c)) \mod c \\ 
(a * b) \mod c \equiv ((a \mod c) * (b \mod c)) \mod c $$
This lets you rewrite your required condition as:
$$((n \mod 9)^{3} + ((n \mod 9) + 1)^{3} + ((n \mod 9) + 2)^{3}) \mod 9 = 0$$
for all $n$, which makes it sufficient to check for $n = 0, 1, \dots, 8$.
A: This is because 


*

*every number is congruent to exactly one of $0,1,\dots ,8$.

*congruences are compatible with addition and multiplication, i.e. if $x\equiv x'$ and $y\equiv y'\mod m$, then $x+y\equiv x'+y'$ and $xy\equiv x'y'\mod m$.

A: Alternatively:
$n^3+(n+1)^3+(n+2)^3$
$=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8$
$=3n^3+9n^2+15n+9$
$=9(n^2+1)+3n(n^2+5)$
In fact, you only need to check for $n=1;n=2;n=3$. 


*

*If $n=1$ or $n \equiv 1 \pmod{3}$ then $n^2\equiv 1\pmod{3}\Rightarrow3|n^2+5\Rightarrow9|n(n^2+5)$

*If $n=2$ or $n \equiv 2 \pmod{3}$ then $n^2\equiv 1\pmod{3}\Rightarrow3|n^2+5\Rightarrow9|n(n^2+5)$

*If $n=3$ or $3|n$ then $3|n^2+5\Rightarrow9|n(n^2+5)$
Note that $9|9(n^2+1)$, this finishes the proof. 
A: Let $n=m-1$
$$(m-1)^3+m^3+(m+1)^3=3m^3+6m=3\underbrace{(m-1)m(m+1)}_{\text{ product of three consecutive integers}}+9m$$
Use The product of $n$ consecutive integers is divisible by $n$ factorial
