There is a single sequence of integers $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, $a_7$ such that \[\frac{5}{7} = \frac{a_2}{2! } + \frac{a_3}{3! } + \frac{a_4}{4! } + \frac{a_5}{5! } + \frac{a_6}{6! } + \frac{a_7}{7! },\] and $0 \le a_i < i$ for $i = 2$, 3, $\dots$, 7. find $a_2 + a_3 + a_4 + a_5 + a_6 + a_7$.

You have a single sequence of integers such that

where for

1. Multiply by 7! to get

By Wilson's theorem,

2. Then write to the left and divide through by 7 to obtain

Repeat this procedure by

And so on:

step-by-step explanation:

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