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Factorial problem

  1. Jan 3, 2016 #1
    What do you think is the value of
    100!-101!+102!-103!.........-109!+110!
    :biggrin:
     
  2. jcsd
  3. Jan 3, 2016 #2

    QuantumQuest

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    n! = (n-1)!n. Take 100! as a common factor and go from there...
     
  4. Jan 3, 2016 #3
    Come on man!!! i tried ... doesn't work....
     
  5. Jan 3, 2016 #4

    Mark44

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    Show us that you tried...
     
  6. Jan 3, 2016 #5
    when you take out 100! common out......then you are left with (1-101+102 times 101...........) which afterwards........don't know man......addition is not the way out as it gives you a very humongous number....
     
  7. Jan 3, 2016 #6

    SteamKing

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    It's going to be pretty large.

    Write the sum this way:

    S = 110! - 109! + 108! - 107! + 106! - 105! + 104! - 103! + 102! - 101! + 100!

    which can be grouped:

    S = (110! - 109!) + (108! - 107!) + ... + (102! - 101!) + 100!

    Now, take the term (110! - 109!) = (110 * 109! - 109!) = (110 - 1) * 109! = 109 * 109!

    You can telescope the other terms in this sum in a similar fashion.

    S = 109 * 109! + 107 * 107! + 105 * 105! + 103 * 103! + 101 * 101! + 100!

    You can manipulate the terms in the sum above in a similar manner, but the result is clear:

    S is a pretty big number no matter how you slice it.

    Were you thinking that S would not be such a large number?
     
  8. Jan 3, 2016 #7
    Does it help any if you take out 110! as a factor?
     
  9. Jan 3, 2016 #8

    fresh_42

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    Are all 179 digits required?
     
  10. Jan 3, 2016 #9

    micromass

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    17038855571963704692695290461249778228462303133623533009426911791940783815733361939707507950770908256181833575228292258746464777211982419630317448315535360000000000000000000000000
     
  11. Jan 3, 2016 #10

    fresh_42

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    Wow! And I was tempted to answer simply O(1). However, Stirling gave me 1.58...for 110! Would be interesting to know whether the calc.exe isn't precise enough or the margin in Stirling's formula is larger than I thought.
     
  12. Jan 3, 2016 #11

    micromass

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    According to the program I just wrote:
    110!=15882455415227429404253703127090772871724410234473563207581748318444567162948183030959960131517678520479243672638179990208521148623422266876757623911219200000000000000000000000000
    So Stirling definitely is accurate here.
     
  13. Jan 3, 2016 #12

    SteamKing

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    Not sure what this number is.
     
  14. Jan 4, 2016 #13

    DrClaude

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    Using Mathematica:
    In[2]:= 110! - 109! + 108! - 107! + 106! - 105! + 104! - 103! + 102! - 101! + 100!

    Out[2]= 15739381947081460468710896569033260448048487750802968746988405111340773775128510600810783940010370922688077274739713895911222137779156961431310006359162880000000000000000000000000
    You can do it yourself using WolframAlpha.
     
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