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


1) 3
2) 4
3) 6
4) 7

• ans 3
  2,3 and 5
• 10 years agoHelpfull: Yes(5) No(1)
• can any one explain it please i didn't get it
• 10 years agoHelpfull: Yes(2) No(1)
• 216=2^3*3^3
  2500=5^4*2^2
  300=2^2*5^2
  after minimizing we get 2^(9/5+2/5+4/5)*3^(9/5+1/5)*5^(8/5+2/5)
  =2^3*3^2*5^2
  =2,3,5
• 9 years agoHelpfull: Yes(2) No(1)
• (6^3)^3/5 * (5^2 * 10^2)^2/5 *(3*10^2)^1/5
  (2^3 * 3^3)^3⁄5 * (5^4 * 2^2)^2/5 * (3* 5^2* 2^2)^1/5
  .'. (2)^ 9/5+4/5+2/5 * (3)^9/5+1/5 * (5)^8/5+2/5
  2^3 * 3^2 * 5^2
  Now, add power of numbers
  3+2+2=7
  therefore, number of prime factor is 7
  ans. option 4) 7
• 3 years agoHelpfull: Yes(1) No(0)
• 3 prime factors are(2,3,5)
• 10 years agoHelpfull: Yes(0) No(0)
• answer is 1)3
• 10 years agoHelpfull: Yes(0) No(0)
• ans 3
  2,3,5
• 9 years agoHelpfull: Yes(0) No(0)
• it can be easily solved as:
  [(6^3)^(3/5)]*[(50^2)^(2/5)]*(30*10)^(1/5)
  after solving it we will get->[(2*3)^5]*[(5*10)^5]*[(2^3*5^2)^5]
  and we can see that the prime factors in the above equation->2,3,5
  Ans:3
• 9 years agoHelpfull: Yes(0) No(0)
• the answer is:7
• 5 years agoHelpfull: Yes(0) No(0)