smc

First, remove all the 90s, since they make no impact. So, we have numbers from $1$ to $10$. Then, $5$ is the 7th number. Let the sum of the first $6$ numbers be $k$. Then, $k\equiv 0\mathrm{mod}6$ and $k\equiv 3\mathrm{mod}7$. We easily solve this as $k\equiv 24\mathrm{mod}42$. Clearly, the sum of the first $6$ numbers must be $570$. In order for the first $5$ numbers to sum to a multiple of $5$, the sixth number must also be a multiple of $5$, since $30$ is a multiple of $5$. Thus, the only option is the sixth number is $10$, which gives $10+90=\boxed{100}$