## h0pe one year ago 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 $$a_2 + a_3 + a_4 + a_5 + a_6 + a_7$$.

1. ganeshie8

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

2. ganeshie8

   puts a new line at the start and end of expression

3. h0pe

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

4. ganeshie8

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

5. freckles

* (bookmarking for tomorrow; sounds interesting to me)

6. h0pe

Thanks @ganeshie8 it looks so much cleaner

7. h0pe

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

8. ganeshie8

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

9. ganeshie8

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

10. h0pe

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

11. ganeshie8

doesn't look correct

12. ganeshie8

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 $$a_5$$

13. h0pe

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

14. ganeshie8

try again

15. ganeshie8

i see lot of typoes/mistakes in ur reply

16. h0pe

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

17. h0pe

is that right?

18. ganeshie8

thats right, but can $$a_5$$ be 5 ?

19. ganeshie8

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

20. h0pe

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

21. ganeshie8

Yes, try working others

22. h0pe

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

23. freckles

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

24. h0pe

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

25. freckles

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