How to prove that a six-digit number of the form $abcabc$ is divisible by 3 distinct primes 
$a,b,c \in \{0,1,2,\ldots,9\}$ with at least one of $a$, $b$, $c$ nonzero. Prove that the six-digit integer $abcabc$ is divisble by at least 3 distinct primes.

I received an answer from the back of the textbook which gave me 
$$abcabc= abc(1001) = (abc)(7)(11)(13),$$
which I am assuming means the 3 distinct primes are 7, 11, 13.
The part I am confused at is that 11, and 13, are not part of the set. Nor does any number in the set $\{0,1,2,\ldots,9\}$ in that a product of the 3 digits would not = 1001. Where did 1001 come from? 
Note: Could I be looking at the question wrong? Where only one of the numbers from the set had to be in $abc$. Where '$a$' could be any of the first 9, while '$b$' and '$c$' could be any number at all? 
 A: It's a little fact that $7 \cdot 11 \cdot 13 = 1001$. And multiplying any three digit number by $1001$ will give a number of the form $abcabc$.
A: \begin{align*}
``abcabc" 
&= 100000a+10000b+1000c+100a+10b+c \\
&= 100100a+10010b+1001c= 1001(100a+10b+c) \\
&= 1001 ⋅ ``abc" \\
1001&= 7 ⋅ 11 ⋅ 13
\end{align*}
A: Nowhere in the question does it specify a set from which the primes must be drawn - you're inserting that yourself. The only requirement the problem imposes on the primes is that they be "distinct" - which just means they all have to be different from one another, so if $abcabc$ happens to be divisible by 4, you can't count 2 twice.
What might be leading you astray is that the problem does specify that $a $, $b $, and $c $ be drawn from 0 through 9. But those are the digits of the number, not the primes. The primes are not given variables in the statement of the problem.
EDIT: It also occurs to me that you may be misinterpreting the notation used in $abcabc$. Importantly, this is not multiplication - otherwise it would have said $a^2b^2c^2$. Instead, it's concatenation - just stick the digits next to each other. For example, if $a=1$, $b=2$, and $c=3$, then $abcabc $ is just the number $123123$, not $36$.
