Question 25
There are unique integers $a_2,$ $a_3,$ $a_4,$ $a_5,$ $a_6,$ $a_7$ such that
$$
\begin{aligned} &\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} \\ &\quad {}+ \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}, \end{aligned}
$$
where $0 \le a_i \lt i$ for $i = 2,$ $3,$ $\ldots,$ $7.$ Find $a_2 + a_3 + a_4 + a_5 + a_6 + a_7.$
$$
\begin{aligned} &\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} \\ &\quad {}+ \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}, \end{aligned}
$$
where $0 \le a_i \lt i$ for $i = 2,$ $3,$ $\ldots,$ $7.$ Find $a_2 + a_3 + a_4 + a_5 + a_6 + a_7.$