if N is a natural no and N^3 has 16 factors then how many maximum factors can N^4 have??a)21 b)24 c)25 d)26

• lets say a term 6
  Now 6^3=(2^3 )* (3^3)
  so, no of factors=(3+1)*(3+1)=16(using prime factorization)
  Now similarly
  6^4=(2^4)*(3^4)
  so factors, (4+1)*(4+1)= 25
  So the ans is c)25
• 11 years agoHelpfull: Yes(50) No(7)
• Consider n = a x b, where a and b are 2 prime factors of n.
  Hence, n^3 = a^3 x b^3. In this case, n^3 will have a total of 4 x 4 = 16 factors.
  Note: The number of factors of a number which can be represented as N = a^p x b^q x c^r , where a, b, c are prime numbers is (p+1)(q+1)(r+1)
  Hence, the number n can be represented only as a product of two prime factors.
  Similarly, n^4 = a^4 x b^4. In this case n^4 will have 5 x 5 = 25 factors.
  
• 10 years agoHelpfull: Yes(10) No(0)
• let....n^3= x^3*y^3........then no/ of factor = (3+1)(3+1)is 16 or possible case
  n^4 = x^4*y^4 = (4+1)(4+1)=25
  one case can also be possible for n^3 let 16 having factor like 2*8 but i think this is impossible case..so only 4*4 case is possible for n^3 having 16 factor.....
• 11 years agoHelpfull: Yes(9) No(6)
• how????? can u plz explain??
• 11 years agoHelpfull: Yes(6) No(4)
• if n=32,ans is 21........ if n=6 ans is 25,,,,,,ambiguity is there
• 8 years agoHelpfull: Yes(3) No(2)
• N^4 can have 25 factors.
• 11 years agoHelpfull: Yes(2) No(4)
• n^y=(y+1)^×
  4^x=16
  x=2
  (4+1)^2=25
• 7 years agoHelpfull: Yes(0) No(0)
• Total factors of a number N=
  a
  p
  .
  b
  q
  .
  c
  r
  .
  .
  .
  is (p+1)(q+1)(r+1)...
  As
  n
  3
  has 16 factors
  n
  3
  can be one of the two formats given below
  n
  3
  =
  a
  15
  
  n
  3
  =
  a
  3
  .
  b
  3
  
  If
  n
  3
  =
  a
  15
  then n =
  a
  5
  and number of factors of
  n
  4
  = 21
  n
  3
  =
  a
  3
  .
  b
  3
  then n = ab and number of factors
  n
  4
  = 25
• 4 years agoHelpfull: Yes(0) No(0)