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 \(i = 2, 3, \ldots, 7.\) Find \(a_2 + a_3 + a_4 + a_5 + a_6 + a_7.\)

I have been stuck on this problem for awhile.

So I figured that in order to solve this problem, I need to get all the denominators the same value. I have been puzzled on how to achieve ths however.

WhoaThere
May 31, 2017

#1**+1 **

The first thing we do is move the a_{7}/7! to the other side:

\(\frac57-\frac{a_7}{7!}=\frac{5\cdot6!}{7!}-\frac{a_7}{7!}=\frac{5\cdot6!-a_7}{7}\frac1{6!},\)

we require the first fraction to be an integer. This is the case for a_{7}=2, now we can bring the next fraction to the left:

\(\frac{514}{6!}-\frac{a_6}{6!}=\frac{514-a_6}{6}\frac{1}{5!},\)

for the first fraction to be an integer a_{6} needs to equal 4. We bring the next fraction to the left:

\(\frac{85}{5!}-\frac{a_5}{5!}=\frac{85-a_5}{5}\frac1{4!},\)

which requires a_{5} to be zero. We bring the next fraction to the left:

\(\frac{17}{4!}-\frac{a_4}{4!}=\frac{17-a_4}{4}\frac1{3!},\)

which requires a_{4} to be 1. We continue on our course:

\(\frac{4}{3!}-\frac{a_3}{3!}=\frac{4-a_3}3\frac12,\)

obviously a_{3} should be one. We are left with:

\(\frac12=\frac{a_2}{2!},\)

and we get a_{2}=1, case solved. Checking our answer with the calculator:

\(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{4}{6!}+\frac{2}{7!}-\frac57=0.0000000000000000143,\)

which is purely a machine error, since 1/7! equals 0.0001984126984127 which is far larger than our error.

Honga
May 31, 2017

#2**+1 **

Note that 5/7 = 0.7142857 [ the 7142857 is a repeating string ]

Try this :

a_{2} = 1, a_{3} = 1, a_{4} = 1, a_{5} = 0, a_{6} = 4, a_{7} = 2

1/2! + 1/3! + 1/4! + 0/5! + 4/6! + 2/7! = 5 / 7

I didn''t really have a mathematical way to solve this.....but, note that

1/2! + 1/3! + 1/4! + 1/5! = 43/60 > 5/7

And

1/2! + 1/3! + 1/4! = 17/24 < 5/7

So......we need that 5/7 - 17/24 = 1/168 = a_{6}/6! + a_{7}/7!

So........it was just a case of letting a_{5} = 0 and then manipulating a_{6} and a_{7} to find a solution

Perhaps some other mathematician can find a more elegant way to the solution....!!!!

CPhill
May 31, 2017