What is the number of odd factors of 49000000?

• number raised in prime factors =2^6 * 5^6 * 7^2
  neglect even base (power of 2)for number of odd factors
  then number of odd factor=7*3(power of 5 raised by 1 * power of 7 raised by 1)=21
• 10 years agoHelpfull: Yes(17) No(0)
• 49000000= 2^6*5^6*7^2
  sum of factors = (2^0+2^1+2^2+2^3+2^4+2^5+2^6)(5^0+5^1+5^2+5^3+5^4+5^5+5^6)(7^0+7^1+7^2)
  no of factor = (7)(7)(3)= 147
  no of odd factor = (1)(7)(3)=21.(bcoz if multiply by factor of 2 number will become even..)
  
  
• 10 years agoHelpfull: Yes(6) No(2)
• 49*1000000
  7^2*5*200000
  7^2*5^2*40000
  7^2*5^3*8000
  7^2*5^4*1600
  7^2*5^5*320
  7^2*5^6*64
  7^2*5^6*2^6
  then odd no are 7$5
  then (2+1)*(7+1)
  21
  
• 9 years agoHelpfull: Yes(3) No(0)
• 49000000 = 7^2*(10^6)
  = 7^2*(5*2)^6
  = 2^6*(7^2*5^6)
  No.of odd factors = 1*(2+1)*(6+1) [since 2^0 = 1]
  = 1*3*7
  =21
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• Saurabh can u explain me how it is solved ...i dint get the solution .
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• 2^0 is valid but other will become even
  hence 3*7=21
  
• 10 years agoHelpfull: Yes(0) No(0)
• 49000000=7^2*1000000
  7^2*10^6
  7^2*(5*2)^6
  7^2*5^6*10^6
  since 5 and 7 are odd number
  so,using formula of finding factors we have
  (2+1)(6+1)=21
• 9 years agoHelpfull: Yes(0) No(0)