How many divisors(including 1,but excluding 1000)are there for the number 1000

• 1000=(2^3)*(5^3)
  So the number of factors will be (3+1)*(3+1)=16 (including 1 & 1000)
  excluding 1000=16-1=15 factors
  
• 11 years agoHelpfull: Yes(65) No(1)
• 15 divisors
  500*2
  250*4
  200*5
  125*8
  100*10
  50*20
  40*25
  1000*1
  
  excluding 1000, we get "15"
• 11 years agoHelpfull: Yes(14) No(5)
• no. of devisors including 1000 are 16
  no. of devisors excluding 1000 are 15
• 11 years agoHelpfull: Yes(3) No(3)
• 1000=10*10*10*10=2*2*2*2=16,16-1=15
• 11 years agoHelpfull: Yes(2) No(10)
• 1*2^3*5^3=1000
  total= 1*(3+1)*(3*1)=16
• 11 years agoHelpfull: Yes(2) No(9)
• 1000/2
  500/2
  250/5
  50/5
  5
  2^3*5^3=3*5=15
• 11 years agoHelpfull: Yes(2) No(1)
• 1000=(2^3)*(5^3)
  So (3+1)*(3+1)=16 (including 1 & 1000)
  excluding 1000=16-1=15 factors
  
• 9 years agoHelpfull: Yes(2) No(0)
• ans is 15
  
  
  
• 11 years agoHelpfull: Yes(1) No(2)
• 1000=2^3*5^3
  (3+1)*(3+1)=16
  therefore answer is 16
  
  also mention the question
  excluding 1000 so 16 -1 = 15
• 9 years agoHelpfull: Yes(1) No(0)
• If the prime factorization of a number is in the form of ap × bq × cr ,
  
  then total number of factors = (p+1) × (q+1) × (r+1)
  
  Prime factorizing 1000 = 23 × 53
  
  ⇒ Total No. of factors of 1000 = (3 + 1) × (3 + 1) = 16
  
  But excluding 1000
  
  ⇒ 16 -1 = 15
• 6 years agoHelpfull: Yes(1) No(0)
• 1000=2^3*5^3
  (3+1)*(3+1)=16
  therefore answer is 16
• 11 years agoHelpfull: Yes(0) No(7)
• answer is 16
• 11 years agoHelpfull: Yes(0) No(9)
• Can you please explain any body clearly like i have to apply that formula for another numbers also
• 9 years agoHelpfull: Yes(0) No(0)
• 1000/2 = 500
  500/2 = 250
  250/2 = 125
  125/5 = 25
  25/5 = 5
  5/5 =1
  2^3 *5^3
  'n' after primefactorization we get
  n = (2^a)×(3^b)×(5^c)×(7^d)......
  
  number of divisors for
  n = (a+1)×(b+1) ×(c+1)........
  soooo.....
  (3+1)(3+1) = 16
  1000 exclude here
  16-1 = 15
  Ans =15
• 7 years agoHelpfull: Yes(0) No(0)