Number of Odd divisor of 9,000,000
a) 21 b) 73 c )74

• 9,000,000=2^6*3^2*5^6
  odd divisors=(2+1)*(6+1)=21
• 9 years agoHelpfull: Yes(33) No(1)
• 9,000,000=2^6*3^2*5^6
  so No of odd divisors=
  for 2= 2^0
  for 5= 5^0,5^1,5^2,5^3,5^4,5^5,5^6
  for 3=3^0,3^1,3^2
  
  so No of divisors=1*7*3
  =21 Ans...
• 9 years agoHelpfull: Yes(7) No(0)
• ans is 21 , same was asked in october 28.
  
• 9 years agoHelpfull: Yes(4) No(0)
• 9,000,000=9*1,000,000
  =3*3*10*10*10*10*10*10
  odd number=3*3*5*5*5*5*5*5
  so 2+6+6+6+1=21
• 9 years agoHelpfull: Yes(3) No(2)
• 9000000=2^6*3^2*5^6
  we find odd diviser then evem power is =1
  2^0*(2+1)*(6+1)=21
• 9 years agoHelpfull: Yes(0) No(0)
• 9000000=3^2*10^6=3^2*2^6*5^6
  odd divisors=(2+1)(0+1)(6+1) we will get odd divisors when power of 2 should be zero.
  so, 3*1*7=21
• 7 years agoHelpfull: Yes(0) No(0)
• answer (a) is the correct answer
  the factor of the 9,000,000 is =3^2 * 2^6 * 5^6
  so total no of the factor is =(2+1)(6+1)(6+1)=147
  so total no of even factor is =(2)(6+1)(6+1)=126
  so total no of odd factor is =total factor-even factor
  =147-126
  =21
• 6 years agoHelpfull: Yes(0) No(0)