Number of prime factors in (216) 3â??5 x (2500) 2â??5 x (300) 1â??5 is :

• (216) 3a??5= 2^3 * 3^4 * 5^1 * a^1 * ?^2
  (2500)2a??5= 2^3 * 5^5 * a^1 * ?^2
  (300) 1a??5= 2^2 * 3^1 * 5^3 * a^1 * ?^2
  
  add all the powers of 2,3,5,a and ? = 2^8 * 3^5 * 5^9 * a^3 * ?^6
  then number of factors = (8+1)(5+1)(9+1)(3+1)(6+1)=9*6*10*4*7
  = 15120
• 10 years agoHelpfull: Yes(10) No(0)
• there 3 prime factors such as 2,3,5
• 10 years agoHelpfull: Yes(2) No(0)
• (216) 3a??5= 2^3*3^5*5^2*a*??
  (2500)2a??5= 2^3*5^5*a*??
  (300)1a??5= 2^2*5^3*3^1*a*??
  
  add all the powers of 2,3,5,a and ?= 2^8*3^6*5^10*a^3*?^6
  then number of factors = (8+1)(6+1)(10+1)(3+1)(6+1)
  = 19404
• 10 years agoHelpfull: Yes(0) No(0)
• Answer is 3
  If you break down each of the three terms 216 has factors 2 and 3
  2500 powers of 5 and 2
  300 powers of 5, 2 and 3
  Finally, only three prime factors are left i.e., 2, 3 and 5
  Note: If you are trying to multiply 1 added to powers of each prime factor, you end up in composite factors.
• 4 years agoHelpfull: Yes(0) No(0)