A number has exactly 3 prime factors, 125 factors of this no. Are perfect square and 27 factors of this no. Are perfect cubes. Overall how many factors does the number have ?

• 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
• 10 years agoHelpfull: Yes(10) No(2)
• (3^125)x(2^57)
• 10 years agoHelpfull: Yes(5) No(1)
• If a number has n prime factors , number of factors that are perfect squares = a^n and number of factors that are perfect cubes = b^n then
  number of factors = based on following conditions
  
  If 2(a-1) > 3(b-1) , then (2a-1)^n
  If 2(a-1) < 3(b-1) , then (2a)^n
  If 2(a-1) = 3(b-1) , then both are possible.
  here no of prime factors=n=3
  so, a^3=125 ; a=5 ; 2(a-1)=8 and
  b^3=27 ; b=3 ;3(b-1)=6
  therefore 2(a-1)> 3(b-1)
  so as per the condition mention above
  the total no of factors=(2a-1)^3=[(2*5)-1]^3=9^3=729.
  
  
• 10 years agoHelpfull: Yes(5) No(0)
• all answers are different .Please provide a correct solution.
• 10 years agoHelpfull: Yes(4) No(0)
• 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
  
• 10 years agoHelpfull: Yes(4) No(0)
• 334
  
  prime factors=3
  squares(x^2=x*x)=125=> 125+125=250
  cubes(x^3=x*x*x)=27 => 27+27+27=81
  
  total factors = 3+250+81 = 334
• 10 years agoHelpfull: Yes(0) No(12)
• a number has aform like a^p *b^q*c^r and so on den the total no of factors for this would be (p+1)(q+1)(r+1) nw in dis question no has 125 perfect sqr that is 125 a^p so total factor from dis is 125*(p+1) here p=2 n coms out to b 125* 3=375 an for cube v have 27*4=108 total will b 483
• 10 years agoHelpfull: Yes(0) No(3)
• let the number is n
  now n=a*b*c* a1^2*a2^2........*a125^2 *b1^3*b2^3........b27^3
  where a,b,c are for three prime numbers.
  a1,a2,a3.......a125 for perfect squares.
  b1,b2,b3........b27 for perfect cubes.
  now so thats why total number of factors are 2^3* 3^125* 4^27
• 10 years agoHelpfull: Yes(0) No(0)