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How to solve factorial related problems

  1. Feb 2, 2012 #1
    Hi all,
    I know what factorials are, obviously, and permutations and combinations, but what I Don't know is, given a problem with factorials in it, is there a general format for solving?
    For instance, an old AMC 10 problem:

    10B-#11. What is the tens digit in the sum 7! + 8! + 9! +
    . . . + 2006! ?
    (A) 1 (B) 3 (C) 4 (D) 6 (E) 9

    How would one go about solving this?
    All help appreciated!
  2. jcsd
  3. Feb 2, 2012 #2


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    Science Advisor

    Hey Thundagere and welcome to the forums.

    This reminds me of a number theory problem.

    For finding the final digit of the number (i.e. the lowest ranked digit in the number), then this reduces to finding the number N (mod 10).

    Now based on this as well as other congruence arithmetic identities, do you have any new ideas that you can use to solve this?
  4. Feb 2, 2012 #3

    Char. Limit

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    Gold Member

    Isn't every factorial above a certain number a multiple of 100, and therefore wouldn't add to the tens digit?
  5. Feb 2, 2012 #4


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    Science Advisor

    That's a great observation and makes the problem really easy :)
  6. Feb 2, 2012 #5
    So anything greater than or equal to 10 we can discount, since 2 * 5 * 10 is 100, which wouldn't matter.
    So with that in mind, we're looking at
    7! + 8! + 9!
    =7!(1 + 8 + 8 * 9)
    So it's then 4?
    For these types of AMC 10 problems, could reasoning it out as above suffice? I'm not well versed in number theory, so....yeah.
    Thanks for your help!
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