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# Number Theory

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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

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

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
edited by WhoaThere  May 31, 2017
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#1
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The first thing we do is move the a7/7! 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 a7=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 a6 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 a5 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 a4 to be 1. We continue on our course:

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

obviously a3 should be one. We are left with:

$$\frac12=\frac{a_2}{2!},$$

and we get a2=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
edited by Honga  May 31, 2017
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Note that   5/7   =  0.7142857     [ the  7142857 is a repeating string ]

Try this  :

a2 = 1, a3  = 1, a4  = 1,  a5  = 0, a6 = 4, a7 = 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   =   a6/6! + a7/7!

So........it was just a case of  letting a5 = 0 and then  manipulating a6  and a7  to find a solution

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

CPhill  May 31, 2017
edited by CPhill  May 31, 2017

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