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

`(a_2)/(2!) + (a_3)/(3!) + (a_4)/(4!) + (a_5)/(5!) + (a_6)/(6!) + (a_7)/(7!)=(5)/(7),`

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

1. Multiply by 7! to get

`(7!a_2)/(2!) + (7!a_3)/(3!) + (7!a_4)/(4!) + (7!a_5)/(5!) + (7!a_6)/(6!) + (7!a_7)/(7!)=(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=(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=(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=(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=(17-1)/(4)=4,\\ \\a_3+3\cdot a_2\equiv 1(\mod 3)\Rightarrow a_3=1,\\ \\a_2=(4-1)/(3)=1.`

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