$\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}$

Question

There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that $$\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}$$where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7.$ Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}.$

Answer 1

Hint: $7!\times\frac{5}{7}=7\times(\frac{6!}{2!}a_2+...+\frac{6!}{6!}a_2)+a_7$, so mod 7, we get $a_7=2$.

Repeatedly apply this procedure, we get $a_6=4, a_5=0, a_4=1, a_3=1, a_2=1$.

As a side note, $$1=\sum_{n=2}^\infty \frac{n-1}{n!}$$ so we immediately get $$1-\frac{5}{7}=\frac{0}{2!}+\frac{1}{3!}+\frac{2}{4!}+\frac{4}{5!}+\frac{1}{6!}+\frac{4}{7!}+\sum_{n=8}^\infty \frac{n-1}{n!}$$