A number has exactly 3 prime factors, 125 factors of this number are perfect squares and 27 factors of this number are perfect cubes. overall how many factors does the number have???

• As it is given no has three prime factors....
  so we may consider it as N=(A^p)(B^q)(C^r)
  now total no of factors of this no is
  (p+1)(q+1)(r+1).
  and no of square factors =([p/2] +1)([q/2] +1)([r/2] +1)
  and no of cubic factors =([p]/3 +1)([q]/3 +1)([r]/3 +1)
  but it is given that the no has 125 square factors i.e
  ([p/2] +1)([q/2] +1)([r/2] +1)=125
  => [p/2]=4.....[r/2]=4
  => p=8,q=8 nd r=8
  similarly for cubic factor
  ([p/3] +1)([q/3] +1)([r/3] +1)=27
  [p/3]+1=3,[q/3]+1=3 nd [r/2]+1=3
  => [p/3]=2=[q/3]=[r/3]
  possible values of p,q,r will be 6,7 or 8
  now total no of factors =(8+1)(8+1)(8+1)=729
  hence answer is 729
  Note:here [X ] means greatest intigr value , either equal to or less than X.
• 12 years agoHelpfull: Yes(52) No(6)
• Answer for this question is 729
  the explanation goes like this
  Let a,b,c be three prime no.
  since the no. of perfect square as a factor is 125
  which can only be if every prime no. has power in the multiple of 2 including zeros(bcoz any no. to the power of zero is a perfect square)
  so power must be 0,2,4,6,8
  thereby 5*5*5=125
  Since no. of cube as a factor is 27
  therefore power in the prime no. will be in the multiple of 3
  so power are 0,3,6
  hence the no. N=a^8*b^8*c^8
  hence total no. of factor is (8+1)(8+1)(8+1)=729
• 12 years agoHelpfull: Yes(22) No(37)
• unable to understand
  
• 12 years agoHelpfull: Yes(17) No(2)
• 125*2=250
  27*3=81
  3 prime factor= 5 factor 1 and no it self
  =250+81+5
  336
• 12 years agoHelpfull: Yes(4) No(23)
• hey akash i am unable to understand your logic can u be more clear
• 12 years agoHelpfull: Yes(2) No(1)
• send that ans clearly
  
• 12 years agoHelpfull: Yes(0) No(2)
• kck plz xplain me the solution..
  
• 9 years agoHelpfull: Yes(0) No(0)