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Sigma math help

  1. Nov 23, 2003 #1
    Problem:

    [tex](n!)^3[/tex] n = {1-99}

    How many digit is the resulting [tex](n!)^3 [/tex]?


    Attempted solution:

    [tex]\Sigma^{99}_{n=1} (n!)^3 = (\Sigma^{99}_{n=1} (n!))^3 = 99(\frac{1!^3 - 99!^3}{2}) =~ 470[/tex] digits. But they say it's wrong. Please help.
     
  2. jcsd
  3. Nov 23, 2003 #2

    Hurkyl

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    [?]

    None of this post made sense... try it again from scratch?
     
  4. Nov 23, 2003 #3
    The problem is:

    (1)!^3 + (2)!^3 + ... + (99)!^3

    How many digits are in the resulting sum?
     
  5. Nov 23, 2003 #4

    Hurkyl

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    Ah, ok, I wasn't sure.

    Anyways, I think it's a trick question. :smile: (99!)^3 is a lot bigger than the rest of the terms in the sequence.

    The formula to compute the number of digits d the number n has is:

    [tex]
    d = \floor{\log_{10} n} + 1
    [/tex]

    And you can use the properties of logarithms to evaluate this when [tex]\mbox{n=(99!)^3}[/tex].


    Once you know how many digits there are in (99!)^3, figure out the maximum possible number of digits in the sum of the rest of them and see if you can prove that the number of digits in (99!)^3 is or is not the number of digits in the sum.

    Hint: For this second part, it may be simpler to first try and solve this brain teaser:
    What is the largest 6 digit number that has the property that if you add a 3 digit number to it, the sum still has 6 digits?
     
  6. Nov 23, 2003 #5
    Hurkl, where did you learn all these "mathematic tricks"? They are very valuable, more like "mathematical treasures!!!!"
     
  7. Nov 23, 2003 #6
    after solving d, the answer came very close to my answer... 469.
     
  8. Nov 23, 2003 #7

    Hurkyl

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    Bah, the floor function didn't appear in my TeX.

    You're supposed to round down in this computation, sorry, so it's 468.


    Basically, I've done a lot of math. :smile: (It's been a hobby ever since I was like 2) The more math you read and do, the more tricks, facts, et cetera you pick up.
     
    Last edited: Nov 23, 2003
  9. Nov 23, 2003 #8
    So was there something wrong with my sigma approach? Rounding?
     
  10. Nov 24, 2003 #9

    NateTG

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    Re: Sigma :)

    Originally posted by PrudensOptimus

    [tex]\Sigma^{99}_{n=1} (n!)^3 = (\Sigma^{99}_{n=1} (n!))^3[/tex]

    Is a pretty huge increase.

    There are a bunch of terms in the middle that you skipped which might account for the extra digits.
     
  11. Nov 24, 2003 #10

    Hurkyl

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    The next step doesn't follow either; it is not true that [tex]\mbox{(\Sigma^{99}_{n=1} (n!))^3 = 99(\frac{1!^3 - 99!^3}{2})}[/tex]
     
  12. Nov 24, 2003 #11
    Isn't it true that:

    [tex]\Sigma^N_{n=1} n_N = N(\frac{n_1 + n_N}{2})[/tex] ?
     
  13. Nov 24, 2003 #12

    NateTG

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    Yeah, but you're dealing with n!, not n.
     
  14. Nov 24, 2003 #13

    Hurkyl

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    What do you mean by [tex]\mbox{n_N}[/tex]?
     
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