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There is a single sequence of integers $a_2,$ $a_3,$ $a_4,$ $a_5,$ $a_6,$ $a_7$ such that
\frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!} = \frac{3}{128}
and $0 \le a_i < i$ for $i = 2,$ $3,$ $\dots,$ $7$.  Find $a_2,$ $a_3,$ $a_4,$ $a_5,$ $a_6,$ $a_7.$

 
 Jul 27, 2024

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