no of factors of 1800?

• N = 1800 => 2^3 * 3^2 * 5^2 [a^x * b^y * c^z ..]
  
  Total no. of factors of N can be calculated by (x+1)*(y+1)*(z+1)
  => (3+1) * (2+1) * (2+1)
  => 36
  
  Ans : 36
• 10 years agoHelpfull: Yes(35) No(1)
• 1800= 3^2*2^3*5^2
  no of factors = (2+1)*(3+1)*(2+1)= 36
• 10 years agoHelpfull: Yes(4) No(0)
• factors for 1800 is 2^3,3^2 and 5^2 so that for no of factor (3+1)*(2+1)*(2+1)=36
• 10 years agoHelpfull: Yes(1) No(0)
• x=1800
  first 2^3*3^2*5^2
  formula a^x*b^y*c^z......
  then (x+1)*(y+1)*(z+1)=(3+1)*(2+1)*(2+1)
  4*3*3=36
  ans 36 factors
• 10 years agoHelpfull: Yes(1) No(0)
• 1800=2^3*3^2*5^2
  no of factors = (3+1)*(2+1)*(2+1)
  =36
• 10 years agoHelpfull: Yes(1) No(1)
• total no of factors=36
• 10 years agoHelpfull: Yes(0) No(1)