find no. of odd divisors n=9000000

• 2^6 *5^6 *3^2
  so odd= 5^6*3^2
  no of factor (6+1)*(2+1)
  =21
  
• 10 years agoHelpfull: Yes(27) No(1)
• given no can be written as 3^2*5^6*2^6
  now we want only odd factors so remove all the powers of two
  which gives us 2^0*3^2*5^6
  now we can get an odd divisor by selecting powers of 2,3 & 5 in (1*3*7) different ways respectively
  which gives us 21 odd factors
• 10 years agoHelpfull: Yes(6) No(0)
• Someone please explain why (6+1)(2+1)??
  
• 10 years agoHelpfull: Yes(4) No(0)
• ans: 21
  
  (6+1)(2+1)=21
• 10 years agoHelpfull: Yes(2) No(0)
• ans :21
  no of terms in the expansion:(2^0)(3^0+3^1+3^2)(5^0+5^1+5^2+5^3+5^4+5^5+5^6)
• 10 years agoHelpfull: Yes(1) No(0)
• 9000000=64*140625
  140625= 5^6*3^2
  (6+1)(2+1)=21
• 10 years agoHelpfull: Yes(1) No(0)
• 9*10^6=3^2*(2*5)^6=3^2*2^6*5^6
  
  
  Now only 3 and 5 can be used as odd factors so we got 3 option for 3 - 0,1,2 and 7 option for 5 0,1,2,3,4,5,6 => 3*7=21
• 10 years agoHelpfull: Yes(1) No(0)
• 21 is abosultely wrong....ans is 73
• 10 years agoHelpfull: Yes(0) No(8)