a number has 3 prime factors, 125 factors are perfect square, 27 factors are perfect cubes,find total number of factors

• We know that the total factors of a number N = ap.bq.cr ....
  Now the total factors which are perfect squares of a number N = ([p2]+1).([q2]+1).([r2]+1)....
  where [x] is greatest intezer less than that of x.
  Given ([p2]+1).([q2]+1).([r2]+1).... = 125
  So [p2]+1 = 5; [q2]+1 = 5; [p2]+1 = 5
  [p2] = 4 ⇒ p = 8 or 9, similarly q = 8 or 9, r = 8 or 9
  Given that 27 factors of this number are perfect cubes
  so ([p3]+1).([q3]+1).([r3]+1).... = 27
  So [p3]+1 = 3 ⇒ = [p3] = 2
  ⇒ p = 6, 7, 8
  By combining we know that p = q = r = 8
  So the given number should be in the format = a8.b8.c8 ....
  Number of factors of this number = (8+1).(8+1).(8+1) = 729
  
• 9 years agoHelpfull: Yes(7) No(2)
• We know that the total factors of a number N = a^p.b^q.c^r ....
  Now the total factors which are perfect squares of a number N = ([p/2]+1).([q/2]+1).([r/2]+1)....
  where [x] is greatest intezer less than that of x.
  Given ([p/2]+1).([q/2]+1).([r/2]+1).... = 125
  So [p/2]+1 = 5; [q/2]+1 = 5; [p/2]+1 = 5
  [p/2] = 4 ⇒ p = 8 or 9, similarly q = 8 or 9, r = 8 or 9
  Given that 27 factors of this number are perfect cubes
  so ([p/3]+1).([q/3]+1).([r/3]+1).... = 27
  So [p/3]+1 = 3 ⇒ = [p/3] = 2
  ⇒ p = 6, 7, 8
  By combining we know that p = q = r = 8
  So the given number should be in the format = a^8.b^8.c^8 ....
  Number of factors of this number = (8+1).(8+1).(8+1) = 729
  
• 8 years agoHelpfull: Yes(2) No(0)
• 729 is the answer
• 9 years agoHelpfull: Yes(0) No(10)