h0pe
  • h0pe
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 \(a_2 + a_3 + a_4 + a_5 + a_6 + a_7\).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
use `\( latex mess goes here \)` for inline latex expressions
ganeshie8
  • ganeshie8
` \[ \] ` puts a new line at the start and end of expression
h0pe
  • h0pe
It's getting late, so I'll fix it when I get on tomorrow.

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More answers

ganeshie8
  • ganeshie8
I fixed it for now, see if it still looks the same
freckles
  • freckles
* (bookmarking for tomorrow; sounds interesting to me)
h0pe
  • h0pe
Thanks @ganeshie8 it looks so much cleaner
h0pe
  • h0pe
I don't understand how you did that...
ganeshie8
  • 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
ganeshie8
  • ganeshie8
i have fixed a typo... please go thru again
h0pe
  • h0pe
For 5 should I now do \[0≡0+a_5+a_6+a_7(mod 5)=a_5=4\] For 4: \[0\equiv0+a_4+a_5+a_6+a_7(mod4)=a_4=2\]
ganeshie8
  • ganeshie8
doesn't look correct
ganeshie8
  • 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 \(a_5\)
h0pe
  • h0pe
Right, forgot that part. \[0\equiv6a_5+37 (\mod5)\] which is \[0\equiv55(\mod5)\] so \[a_5=3\]
ganeshie8
  • ganeshie8
try again
ganeshie8
  • ganeshie8
i see lot of typoes/mistakes in ur reply
h0pe
  • h0pe
So first \[0\equiv7*6a_5+28+2(mod 5)=0\equiv42a_5+30(mod 5)\] Then \[a_5=5\]
h0pe
  • h0pe
is that right?
ganeshie8
  • ganeshie8
thats right, but can \(a_5\) be 5 ?
ganeshie8
  • ganeshie8
from hypothesis \(0\le a_5\lt 5\) right
h0pe
  • h0pe
Oh, right. \(a_5\) has to be 0.
ganeshie8
  • ganeshie8
Yes, try working others
h0pe
  • 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
freckles
  • 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
h0pe
  • h0pe
so complicated ugh I'm going to write this huge thing down on paper
freckles
  • freckles
6!=6*5*4*3*2 ***

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