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

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

ganeshie8 · Answer

use `$ latex mess goes here $` for inline latex expressions

ganeshie8 · Answer

`   ` puts a new line at the start and end of expression

h0pe · Answer

It's getting late, so I'll fix it when I get on tomorrow.

ganeshie8 · Answer

I fixed it for now, see if it still looks the same

freckles · Answer

* (bookmarking for tomorrow; sounds interesting to me)

h0pe · Answer

Thanks @ganeshie8  it looks so much cleaner

h0pe · Answer

I don't understand how you did that...

ganeshie8 · Answer

$$\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!},$$

multuply $7!$ through out and get
$$5\cdot 6! = 7\cdot 6\cdot 5\cdot 4\cdot 3a_2+7\cdot 6\cdot 5\cdot 4a_3+7\cdot 6\cdot 5a_4+7\cdot 6a_5+7a_6+a_7$$

taking $\mod 7$ both sides gives
$$5(-1)\equiv 0+a_7\pmod{7} \implies a_7 = 2$$

taking $\mod 6$ both sides gives
$$0\equiv 0+a_6+a_7\pmod{6} \implies a_6 = 4$$

see if you can find other values similarly

ganeshie8 · Answer

i have fixed a typo... please go thru again

h0pe · Answer

For 5 should I now do $$0≡0+a_5+a_6+a_7(mod 5)=a_5=4$$
For 4: $$0\equiv0+a_4+a_5+a_6+a_7(mod4)=a_4=2$$

ganeshie8 · Answer

doesn't look correct

ganeshie8 · Answer

we have equation : $$5\cdot 6! = 7\cdot 6\cdot 5\cdot 4\cdot 3a_2+7\cdot 6\cdot 5\cdot 4a_3+7\cdot 6\cdot 5a_4+7\cdot 6a_5+7a_6+a_7$$

taking mod5 should give
$$0\equiv 0+7\cdot 6a_5 + 7a_6 + a_7 \pmod{5}$$

plugin the previous known values and solve $a_5$

h0pe · Answer

Right, forgot that part.
$$0\equiv6a_5+37 (\mod5)$$ which is $$0\equiv55(\mod5)$$ so $$a_5=3$$

ganeshie8 · Answer

try again

ganeshie8 · Answer

i see lot of typoes/mistakes in ur reply

h0pe · Answer

So first
$$0\equiv7*6a_5+28+2(mod 5)=0\equiv42a_5+30(mod 5)$$
Then $$a_5=5$$

h0pe · Answer

is that right?

ganeshie8 · Answer

thats right, but can $a_5$ be 5 ?

ganeshie8 · Answer

from hypothesis $0\le a_5\lt 5$ right

h0pe · Answer

Oh, right. $a_5$ has to be 0.

ganeshie8 · Answer

Yes, try working others

h0pe · Answer

$$5⋅6!=7⋅6⋅5⋅4⋅3a_2+7⋅6⋅5⋅4a_3+7⋅6⋅5a_4+7⋅6a_5+7a_6+a_7$$
How do you take mod 4

freckles · Answer

Anything that with a factor 4 will have a remainder of 0 when dividing the thing by 4. 
That is like for example 6!=5*4*3*2
so 6! mod 4 is 0

h0pe · Answer

so complicated ugh I'm going to write this huge thing down on paper

freckles · Answer

6!=6*5*4*3*2 ***