How many factors of the number 2^5*3^4*5^2 are divisible by 2?

• total no. of factors= (5+1)(4+1)(2+1)= 90
  
  no.of odd factors= (4+1)(2+1)= 15
  
  so, no. of even factors= 90-15=75
• 11 years agoHelpfull: Yes(26) No(0)
• 75 is the answer,
  (32*81*25)/2 is 2^4*3^4*5^2..so total no.of factors are (4+1)*(4+1)*(2+1)=75
• 11 years agoHelpfull: Yes(7) No(0)
• for the given number the required factors are
  with atleast one 2 and remaining any thing from the given expression
  2 : 1 combination
  2*(_) 3 or 5 : 2 combinations
  2*(_*_) 3*3 or 3*5 or 5*5: 3combs
  2*(_*_*_) 3*3*3 or 3*3*5 or 3*5*5 :3 combs
  2*(_*_*_*_) 3*3*3*3 or 3*3*3*5 or 3*3*5*5 :3combs
  2*(_*_*_*_*_) 3*3*3*3*5 or 3*3*3*5*5 : 2 combs
  2*(_*_*_*_*_*_) 3*3*3*3*5*5 : 1comb
  
  total 15 combs for 2
  repeating same for 2^2 2^3 2^4 2^5 we get
  15*5=75 factors
  
• 11 years agoHelpfull: Yes(5) No(0)
• ans) would be 19
  
  12 = (2^2)*3
  
  (2^2)*3 this should be multiple of the no.
  
  remaining is 2^3*3^3*5^2
  
  3*3*2=18
  
  total would be 18+1=19
• 11 years agoHelpfull: Yes(0) No(8)
• fcrf jqdfghmmsdfhjkldfgh ryuol;'
  ghjkl. 6yu8i9 yujkl;'
• 10 years agoHelpfull: Yes(0) No(0)