Q). Number of prime factors in (216)^3 / 5 x (2500)^2 / 5 x (300)^1 / 5 is :
Option
1) 3
2) 4
3) 6
4) 7

• 3
  solving it we get 216*3*100*2*60=216*6*6*1000
  =6^5*10^3
  so 2,3,5 are prime factors=3
• 10 years agoHelpfull: Yes(15) No(4)
• 216*3/5*2500*2/5*300*1/5
  dividing by all 3 5
  we get 216*3*100*2*60
  now write it in prime factor format
  2^8* 3^2*5^3*17^1
  so prime factor will be 9*3*4*2=216
  so 216 prime factors :)
• 10 years agoHelpfull: Yes(5) No(13)
• ans is 3????
  
• 10 years agoHelpfull: Yes(2) No(0)
• solving it we get 2^8*3^7*5^3
  so 2,3,and 5 are prime factors
  so ans is 3.
• 10 years agoHelpfull: Yes(2) No(1)
• 3 is ans...
• 10 years agoHelpfull: Yes(2) No(1)
• sorry above solution was wrong...
  after solving we will get the prime factors as 2^8*3^5*5^3
  so number of prime factors will be=9*6*4
  216
• 10 years agoHelpfull: Yes(2) No(4)
• answer 3
  3 factors are 3 ,2 no 5,
• 8 years agoHelpfull: Yes(1) No(0)
• plz clearly xplain the ans
  
• 7 years agoHelpfull: Yes(1) No(0)
• please explain the answer....
• 8 years agoHelpfull: Yes(0) No(0)
• after solving powerr
  2^3*3^2*5^1=7
  
• 8 years agoHelpfull: Yes(0) No(0)