Let $N$ be any $6$-digit positive integer in base 10. Let $N^*$ be obtained from $N$ by swapping the positions of the first $3$ digits with the last $3$ digits. (eg. if $N = 123456$, then $N^* = 456123$). Prove that for any choice of $N$, $N + N^*$ is divisible by 7.
I'm confused as to where to begin this question. Am I supposed to create variables that represent each digit for both $N$ and $N^*$ before starting the proof? We're also learning about congruence so I'm assuming that might be used somewhere within the solution.
If $N=10^3a+b,N^*=10^3b+a$
$\implies N+N^*=(10^3+1)(a+b)$
Now $1001=7\cdot11\cdot13$