how many divisors are there for 1758

• 1758 = 2*3*293
  293 is a prime no.
  so the no. of divisors is given by
  (1+1)*(1+1)*(1+1) = 8
  where the first 1 in the bracket is the power of the factors.
• 9 years agoHelpfull: Yes(13) No(1)
• 2^1 * 3^1 * 293^1
  No of divisors = (1+1)*(1+1)*(1+1)=8
• 9 years agoHelpfull: Yes(2) No(0)
• 8
  soln:
  1758=2*3*293
  (3c1 + 3c2 + 3c3) and 1 is also a divisor,so(3+3+1)+1=8
  
• 9 years agoHelpfull: Yes(1) No(0)
• 1758= 2^6 * 3^3
  so, (6+1)*(3+1) .....{by formula}
  =7*4=28
  28 divisors are there for 1758.
• 9 years agoHelpfull: Yes(1) No(9)
• 1758=2*3*293
  (1+1)*(1+1)*(1+1) = 8
• 6 years agoHelpfull: Yes(1) No(0)