1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Divisors of a number

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data
    let N= (7^n1)(9^n2)(11^n1)
    if n1 is even,then the total number of divisors of the form of 4k+1 must be divisible by???

    correct answer is n2+1

    3. The attempt at a solution
    total number of divisors = (n1+1)(n2+1)(n1+1)
     
    Last edited: Apr 9, 2010
  2. jcsd
  3. Apr 9, 2010 #2
    can anybody plz help me uot in this question????
     
  4. Apr 9, 2010 #3
    can anybody plz help me out in this question????
     
  5. Apr 9, 2010 #4
    Hm... I don't have the solution, but can give some idea: The factor decomposition is funny, using 9 instead of 3. Anyway, put it like 9^n2 (7*11)^n1. Now get numbers of the form 4k+1: 1, 5, 9, 13, 17, 21, 25... 81... So, you see: powers of 9 are of the form 4k+1. You have to prove that no number of that form can be a multiple of 7 or 11.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Divisors of a number
  1. Maximum common divisor (Replies: 5)

Loading...