Find the sum of all the possible proper three digit numbers $\overline{abc}$ such that the six-digit number $\overline{741abc}$ is divisible by $6,7$ and $10$.
In this problem, what's bothering me is the overline over the numbers — do they have any significant meaning? Also, by brute-forcing, I was able to get that $\overline{abc}$ could be $090,300,510,720$ and $930$, but the sum apparently seems to have not counted $090$ into account, and I was wondering about that. (Okay, the biggest hint is that $c=0$ is a must since $10$ divides the number and that takes care of the $2$ in $6$, so I only have to make sure $ab$ is divisible by $3$ and $741ab$ is divisible by $7$.) Even though "a" is technically within a number, do I still have to make $a\neq 0$?
The overline indicates that $abc$ is to be read as three digits, not as the product of $a,b,$ and $c$. When you are asked for a proper number $\overline {abc}$ they mean to exclude $a=0$. $a$ is not in the middle of this one.