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Factors of a number

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data
    N has total 105 factors including 1 and N. then find :
    a) the total no of odd factors between 1 and N.
    b) if the total number of divisors of N which are multiple of 36 are 45.then the total no of odd factors between 1 and N.
    c)the number of ways in which N can be resolved into 2 factors which are relatively prime to each other is equal to 4,then the total no of odd factors between 1 and N.
    d) if the total number of divisors of N which are multiple of 216 are 48,then the total no of odd factors between 1 and N.

    3. The attempt at a solution
    total number of divisors of a number a^n1*b^n2*c^n3 is equal to (n1+1)(n2+1)(n3+1)
     
    Last edited: Apr 9, 2010
  2. jcsd
  3. Apr 9, 2010 #2

    tiny-tim

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    Hi jeedoubts! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)
    Well, 105 = 3*5*7, so how does that help you with a) ? :smile:
     
  4. Apr 9, 2010 #3
    we can assume N to be a2b4c6
    and if check if either of a or b or c is even or not so in all 4 answers are possible.... Ithink in that way plz tell if i'm correct...
     
  5. Apr 9, 2010 #4

    tiny-tim

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    a b and c must be primes, so only one (or zero) of them can be even …

    but there doesn't seem to be enough information to answer a) :confused:
     
  6. Apr 9, 2010 #5

    what about parts b,c and d??:confused::confused:
     
  7. Apr 9, 2010 #6

    tiny-tim

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    Can you check the question?

    Are you sure it doesn't start with 135 (rather than 105) ?
     
  8. Apr 9, 2010 #7
    it is 105 i checked it.
     
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