find the intersection between two sets $A=\{x: x\in {Z} , x$ is sum of 7 consecutive integers $\}$
$B=\{x: x\in {Z} , x$ is sum of 8 consecutive integers 
$C=\{x: x\in {Z} , x$ is sum of 9 consecutive integers $\}$
find $A \cap B \cap C.$
I got this problem from "Challenges and thrills on pre-college mathematics"
my first attempt was 
$A=\{x\in {Z}:x=7n+28,\forall n \in {N} \}$
$B=\{x\in {Z}:x=8n+36,\forall n \in {N} \}$
$C=\{x\in {Z}:x=9n+55,\forall n \in {N} \}$
LCM of $7,8,9$ is $504$.
therefore $A \cap B \cap C = \{504x:\forall x\in {Z} \}$
is it correct?
 A: There is a small error in the description of the sets $A$, $B$ and $C$. For $x\in A$ you can prove that there exists a $n\in\mathbb Z$ such that $x=7n+28$. Then you have to write
$$
A=\{x\in\mathbb Z~:~\exists n\in\mathbb Z\text{ such that }x=7n+28\}.
$$
The restriction $n\in\mathbb N$ is wrong and $\forall$ is wrong too.
But you can do it better. Because of $28=4\cdot 7$ you can say
$$
A=\{x\in\mathbb Z~:~\exists n\in\mathbb Z\text{ such that }x=7n\}.
$$
Further, you get
$$
B=\{x\in\mathbb Z~:~\exists n\in\mathbb Z\text{ such that }x=8n+4\}
$$
$$
C=\{x\in\mathbb Z~:~\exists n\in\mathbb Z\text{ such that }x=9n\}.
$$
From $x\in A\cap B\cap C$ you can conclude that there exists $n\in\mathbb Z$ such that $x=7\cdot(8n+4)\cdot 9=504n+252$. Hence, 
$$
A\cap B\cap C\subset\{x\in\mathbb Z~:~\exists n\in\mathbb Z\text{ such that }x=504n+252\}=:D.
$$
Now, consider $x\in D$ and you have to argue why $x\in A\cap B\cap C$ holds, which is easy to see.
A: I think that there is an error (In the first answer)
$$C=\{x\in\mathbb Z~:~\exists n\in\mathbb Z\text{ such that }x=9n{\color{Red} +1}\}$$
