Looking for some faster ways to solve this Could you show me that how can I solve this as fast as possible, please? 
$$ ABCDEF \, \land \, DEFABC \; \large{ \in \, \mathbb{N^{+}} } $$  $$ ABCDEF \, = \, 6 \times (DEFABC) \\ $$  $$ (A+B+C+D+E+F)=\; ? $$
Thank you very much... :)
 A: Here's another solution: we immediately see that $D=1$, since otherwise $6\times DEFABC$ would have to have more than seven digits. Now write: $X=ABC$, $Y=DEF$. The equation becomes: $1000X+Y=6(1000Y+X)$ which can be rewritten as $994X=5999Y$. Now, since $994$ is divisible by the primes $2$ and $71$, and $5999$ isn't, $Y$ will have to be divisible by $2$ and $71$. But then $Y$ is divisible by $142$ and since the first digit of $Y$ is $1$, it follows that $Y=142$. Now $X=857$ follows easily and we arrive at the same (unique) solution.
A: DEFABC = 142857 obviously here.
More systematically though, let the sum be X. The equation shows that X is divisible 3, so applying it again shows that X is divisible by 9. Assuming that ABCDEF are all distinct, we have to choose 4 digits to reject. Overall sum of 0-9 is 45, so it must be less than that. Sum of rejects being 9 would make it hard to construct DEFABC as 0,1 would have to be rejected. Sum of rejects being 27 would have required us to reject 9, making ABCDEF hard to construct. So sum of rejects must be 18, thus X is 27.
