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Ishaan94

  • 3 years ago

First answer (by Anders) on http://www.quora.com/Mathematics/How-would-I-show-that-49-divides-8-n-7n-1-for-all-n-ge-0 Can anyone explain it to me? Hints'd be appreciated.

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  1. Ishaan94
    • 3 years ago
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    @mukushla @KingGeorge @Zarkon

  2. mukushla
    • 3 years ago
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    can u type the equation here plz :)

  3. Ishaan94
    • 3 years ago
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    @mukushla try again.

  4. Ishaan94
    • 3 years ago
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    I am not concerned with problem as much as its solution on Quora. And No @nincompoop

  5. mukushla
    • 3 years ago
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    thats not a true statement of binomial theorem\[8^n=(1+7)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 7^n\]right?

  6. Ishaan94
    • 3 years ago
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    I am pretty sure, it is.

  7. mukushla
    • 3 years ago
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    oops sorry that should be\[8^n=(1+7)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 7^k\]

  8. Ishaan94
    • 3 years ago
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    oh yeah. k. lol

  9. Ishaan94
    • 3 years ago
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    but that's not what concerns me. it's the first answer by Anders i am unable to understand.

  10. mukushla
    • 3 years ago
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    oh sorry man lol :) im sooo blind

  11. mukushla
    • 3 years ago
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    i cant understand it :)

  12. experimentX
    • 3 years ago
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    \[ 8^n=(1+7)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 7^n = 1 + 7n + O(7^{(n \geq 2)})\]

  13. estudier
    • 3 years ago
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    Sai Ganesh explanation also no good, right?

  14. experimentX
    • 3 years ago
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    \( O(7^{(n \geq 2)}) \) is always divisible by 49

  15. estudier
    • 3 years ago
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    Yes, sorry, I was directing at Ishaan (he wants an "octal" explanation)

  16. Ishaan94
    • 3 years ago
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    Yeah. It didn't help my puny brain much :(

  17. Ishaan94
    • 3 years ago
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    @him1618 maybe you can?

  18. him1618
    • 3 years ago
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    the explanation is good enough

  19. estudier
    • 3 years ago
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    I have to say that the binomial answer is much more natural, it would not occur to me to set out on an octal adventure for that particular problem....

  20. him1618
    • 3 years ago
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    true

  21. Ishaan94
    • 3 years ago
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    Specifically, I don't understand how did he conclude 'The rest can be divided into equal groups based on their first two nonzero digits.'

  22. Ishaan94
    • 3 years ago
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    ... 49* equal groups ...

  23. estudier
    • 3 years ago
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    7*7

  24. estudier
    • 3 years ago
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    you got rid of the 0's so you got 7 digits left you can set, right?

  25. Ishaan94
    • 3 years ago
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    right.

  26. estudier
    • 3 years ago
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    So if all the numbers are divisible into 49 groups, then divisibilty by 49 follows. At least, that's what I think it says.

  27. Ishaan94
    • 3 years ago
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    Okay, one silly doubt. Why is he using only first two digits for grouping?

  28. estudier
    • 3 years ago
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    because 7*7 = 49

  29. Ishaan94
    • 3 years ago
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    i was acting foolishly yesterday :( thanks.

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