A Number has exactly 3 prime factors,125 factors of this number are perfect square and 27 factors of this number are perfect cubes,total how many factors does this number have?

• i think the answer is 729, here goes my logic
  
  N = p1 ^ (x) * p2 ^ (y) * p3 ^ (z)
  
  where p1, p2, p3 are primes and their powers are x, y and z
  
  if i have to form a perfect cube i must choose x, y and z to be multiples of 3 starting from 0, 3, 6, 9, 12, ....
  if i need 27 perfect cubes then each of the powers can take the following values 0, 3, 6 (3 possible values for the 3 powers)
  
  hence we would be looking at N = p1 ^ (6) * p2 ^ (6) * p3 ^ (6)
  these are 27 combinations = 27 perfect cubes
  since there are 3 choices for every power ie. x, y, z the total number of combinations such that i choose a multiple of 3 for x, y, z is 3 * 3 * 3 = 27
  
  Similar logic for the perfect squares
  x, y, z must be multiples of 2 starting from 0, 2, 4, 6, 8, 1,....
  if i need 125 perfect squares the each of the powers can take the following values 0, 2, 4, 6, 8 (5 posible values for the 3 powers)
  hence we would be looking at N = p1 ^ ( * p2 ^ ( * p3 ^ (
  thse are 125 combinations = 125 perfet squares
  
  since this is the larger case which includes the case for perfect cubes i'll asume
  N = p1 ^ ( * p2 ^ ( * p3 ^ (
  
  total number of factors = (8 + 1) * (8 + 1) * (8 + 1) = 9 * 9 * 9 = 729
• 11 years agoHelpfull: Yes(5) No(5)
• After a series of calculations,I got the following shortcut !!!
  
  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
  
  3b = (2a-1) or 2a or (2a+2)
  
  Number of factors of the number = (2a)^n or (2a - 1)^n
  
  Here a = 5 , b = 3 , n = 3
  => number of factors = either 9^3 or 10^3
  = 729 or 1000
  
• 11 years agoHelpfull: Yes(4) No(3)
• number of factors = either (2a)^n or (2a-1)^n 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 2(a-1) > 3(b-1)
  => no of factors = (2a-1)^n = 9^3= 729
• 11 years agoHelpfull: Yes(4) No(2)
• ANS IS 729
• 11 years agoHelpfull: Yes(3) No(2)
• 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-10^3=[(2*5)-1]^3=9^3=729.
  
  
• 10 years agoHelpfull: Yes(3) No(0)
• Thanks Vinay Sipani...d short cut works nicely.....
• 11 years agoHelpfull: Yes(1) No(1)
• is dere any short-cut method 4 dis???
• 11 years agoHelpfull: Yes(0) No(2)