Find the number of divisors of 1728.?

• 1728= 2^6 * 3^3
  Hence the Number of factors = (6+1) x (3+1) = 7 x 4 = 28.
  We know that if a number represented in standard form (a^m *b^n) , then the number of factors Is given by (m+1)(n+1).
  
  Answer is 28
• 9 years agoHelpfull: Yes(78) No(2)
• someone solve the same question for 1248?
• 6 years agoHelpfull: Yes(6) No(0)
• By Prime factorization 1728=2^6*3^3
  
  increasing powers by one and multiplying the powers will generate the answer
  =(6+1)*(3+1)
  =7*4
  =28-answer
• 9 years agoHelpfull: Yes(4) No(0)
• 1728=2^6*3^3
  no.of factors=7*4=28
• 9 years agoHelpfull: Yes(3) No(2)
• 1728=2^6×3^3
  So no. Of divisor of 1728=(6+1)(3+1)=28
• 9 years agoHelpfull: Yes(2) No(2)
• 28
  1728=2^6*3^3
  no of divisors=(6+1)*(3+1)=7*4=28
• 9 years agoHelpfull: Yes(2) No(0)
• i cnt undrstnd plz xplain it clearly
  
• 9 years agoHelpfull: Yes(2) No(2)
• 1728
  =(2^6)*(3^3)
  =(6+1)*(3+1) //Addition of power+1
  =7*4
  =28
• 8 years agoHelpfull: Yes(2) No(0)
• Power of 2 is 6
  and power of 47 is 1
  so (6+1)*(1+1)=14
• 9 years agoHelpfull: Yes(0) No(20)
• lcm of 1728=2^6*3^3
  so no. of divisor =7+4=11
  
• 9 years agoHelpfull: Yes(0) No(12)
• what can we do if the no can be represented in the form of a^b*c^d*e^f...?
• 9 years agoHelpfull: Yes(0) No(1)
• 1728= 2^6 * 3^3
  7 x 4 = 28
  
• 8 years agoHelpfull: Yes(0) No(0)
• 1728 is written in the form of (a^m*a^n)= (2^6*3^3)
  we know that,(a^m*b^n)=(m+1)(n+1)
  so,here m=6 and n=3
  ans:(6+1)(3+1)=7*4
  28
  
• 7 years agoHelpfull: Yes(0) No(0)