The product of 3 natural number (N1,N2,N3) is 12 times their hcf. How many ordered triplets ( N1, N2, N3) are possible?

• Let HCF be H
  N1=Ha
  N2=Hb
  N3=Hc
  Product of 3 numbers=>(H^3)(abc)=12H
  So (H^2)(abc)=12
  In this case H^2 will be either 1 or 4 because if its 9 or more than that then we wont get (H^2)(abc) as 12
  So if H^2 = 1
  then ordered triplets will be
  (1,1,12)(1,2,6)(1,3,4)(1,4,3)(1,6,2)(1,12,1)(2,1,6)(2,2,3)(2,3,2)(2,6,1)(3,1,4)(3,2,2)(3,4,1)(4,1,3)(4,3,1)(6,1,2)(6,2,1)(12,1,1)
  for H^2=4
  The ordered triplets will be :
  (1,1,3)(1,3,1)(3,1,1)
  
  Hence total number of ordered triplets will be 21
• 8 years agoHelpfull: Yes(2) No(0)
• 9 is the ans
• 8 years agoHelpfull: Yes(1) No(0)
• suppose division is 2,2,3
  __ __ __ filling these 3 places with 2,2,3 ways=3!/2!=3
  
  division is 1,3,4
  __ __ __ fill these places with 1,3,4 ways=3!=6
  
  division is 1,2,6
  __ __ __ fill these places with 1,2,6 ways=3!=6
  
  division is 1,1,12
  __ __ __ fill these places with 1,1,12 ways=3!/2!=3
  
  
  
  Ans.18 ordered triplets possible
• 7 years agoHelpfull: Yes(1) No(0)
• how the answer is 9?
  
• 8 years agoHelpfull: Yes(0) No(0)