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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 \(a_2 + a_3 + a_4 + a_5 + a_6 + a_7\).

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  1. ganeshie8
    • one year ago
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    use `\( latex mess goes here \)` for inline latex expressions

  2. ganeshie8
    • one year ago
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    ` \[ \] ` puts a new line at the start and end of expression

  3. h0pe
    • one year ago
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    It's getting late, so I'll fix it when I get on tomorrow.

  4. ganeshie8
    • one year ago
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    I fixed it for now, see if it still looks the same

  5. freckles
    • one year ago
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    * (bookmarking for tomorrow; sounds interesting to me)

  6. h0pe
    • one year ago
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    Thanks @ganeshie8 it looks so much cleaner

  7. h0pe
    • one year ago
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    I don't understand how you did that...

  8. ganeshie8
    • one year ago
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    \[\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!},\] 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
    • one year ago
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    i have fixed a typo... please go thru again

  10. h0pe
    • one year ago
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    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
    • one year ago
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    doesn't look correct

  12. ganeshie8
    • one year ago
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    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
    • one year ago
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    Right, forgot that part. \[0\equiv6a_5+37 (\mod5)\] which is \[0\equiv55(\mod5)\] so \[a_5=3\]

  14. ganeshie8
    • one year ago
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    try again

  15. ganeshie8
    • one year ago
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    i see lot of typoes/mistakes in ur reply

  16. h0pe
    • one year ago
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    So first \[0\equiv7*6a_5+28+2(mod 5)=0\equiv42a_5+30(mod 5)\] Then \[a_5=5\]

  17. h0pe
    • one year ago
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    is that right?

  18. ganeshie8
    • one year ago
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    thats right, but can \(a_5\) be 5 ?

  19. ganeshie8
    • one year ago
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    from hypothesis \(0\le a_5\lt 5\) right

  20. h0pe
    • one year ago
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    Oh, right. \(a_5\) has to be 0.

  21. ganeshie8
    • one year ago
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    Yes, try working others

  22. h0pe
    • one year ago
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    \[5⋅6!=7⋅6⋅5⋅4⋅3a_2+7⋅6⋅5⋅4a_3+7⋅6⋅5a_4+7⋅6a_5+7a_6+a_7\] How do you take mod 4

  23. freckles
    • one year ago
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    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
    • one year ago
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    so complicated ugh I'm going to write this huge thing down on paper

  25. freckles
    • one year ago
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    6!=6*5*4*3*2 ***

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