Question

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 $a_2 + a_3 + a_4 + a_5 + a_6 + a_7$.

Answer 1

You have a single sequence of integers $a_2,\ a_3,\ a_4,\ a_5,\ a_6,\ a_7$ such that

$\dfrac{a_2}{2!} + \dfrac{a_3}{3!} + \dfrac{a_4}{4!} + \dfrac{a_5}{5!} + \dfrac{a_6}{6!} + \dfrac{a_7}{7!}=\dfrac{5}{7},$

where $0 \le a_i < i$ for $i = 2, 3, \dots, 7.$

1. Multiply by 7! to get

$\dfrac{7!a_2}{2!} + \dfrac{7!a_3}{3!} + \dfrac{7!a_4}{4!} + \dfrac{7!a_5}{5!} + \dfrac{7!a_6}{6!} + \dfrac{7!a_7}{7!}=\dfrac{7!\cdot 5}{7},\\ \\7\cdot 6\cdot 5\cdot 4\cdot 3\cdoa a_2+7\cdot 6\cdot 5\cdot 4\cdot a_3+7\cdot 6\cdot 5\cdot a_4+7\cdot 6\cdot a_5+7\cdot a_6+a_7=6!\cdot 5,\\ \\7(6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)+a_7=3600.$

By Wilson's theorem,

$a_7+7\cdot (6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)\equiv 2(\mod 7)\Rightarrow a_7=2.$

2. Then write $a_7$ to the left and divide through by 7 to obtain

$6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6=\dfrac{3600-2}{7}=514.$

Repeat this procedure by $\mod 6:$

$a_6+6(5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5)\equiv 4(\mod 6)\Rightarrow a_6=4.$

And so on:

$5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5=\dfrac{514-4}{6}=85,\\ \\a_5+5(4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4)\equiv 0(\mod 5)\Rightarrow a_5=0,\\ \\4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4=\dfrac{85-0}{5}=17,\\ \\a_4+4(3\cdoa a_2+ a_3)\equiv 1(\mod 4)\Rightarrow a_4=1,\\ \\3\cdoa a_2+ a_3=\dfrac{17-1}{4}=4,\\ \\a_3+3\cdot a_2\equiv 1(\mod 3)\Rightarrow a_3=1,\\ \\a_2=\dfrac{4-1}{3}=1.$

Answer: $a_2=1,\ a_3=1,\ a_4=1,\ a_5=0,\ a_6=4,\ a_7=2.$