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[Proof] Complete square

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Proof that for [tex]n\geq 5[/tex] expression [tex]\sum_{i=1}^{n}i![/tex] can't be a complete square


    3. The attempt at a solution
    Mathematical induction maybe?
    [tex]n=5[/tex] [tex]\sum_{i=1}^{5}i!=1+2!+3!+4!+5!=152[/tex] OK
    [tex]n\rightarrow n+1[/tex] [tex]\sum_{i=1}^{n+1}i!=\sum_{i=1}^{n}i!+(n+1)![/tex]
    ...
    One guy said that
    But, I don't understand him.

    Sorry for bad English.
     
  2. jcsd
  3. Dec 16, 2011 #2
    1+2+6+24+120 = 153 right?
    subsequent factorials are all divisible by 10, so dividing by 5 will always give the remainder 3.
     
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