How many six digits nos. can be formed using digits 1 to 6 such that no. is always divisible by digit at its unit place??

• unit digit=1,no. is always divisible by 1 => 5!
  
  unit digit=2, no. is always divisible by 2 => 5!
  
  unit digit=3, no. is always divisible by 3 => 5! [sum of digits=1+2+3+4+5+6=21,always divisible by 3]
  
  unit digit=4, no. will be divisible by 4 if 10th digit 2 or 6 => 2* 4!
  
  unit digit=5, no. is always divisible by 5 => 5!
  
  unit digit=6, no. is always divisible by 6 => 5! [as, no. ending with 6 is divisible by 2 & sum of digits=21 so divisible by 3 , hence divisible by 6]
  
  total= 5! + 5! + 5! + 2*4! +5! + 5! = 5*5! + 2*4! = 648
  
• 10 years agoHelpfull: Yes(22) No(0)
• hello manikandan, why are you simply copying above answers... rather post your own better explanation... I saw your answers in other posts also... so telling this
• 10 years agoHelpfull: Yes(10) No(0)
• _ _ _ _ _ 1 =>5!=5*4*3*2*1=120
  _ _ _ _ _ 2 =>5!=5*4*3*2*1=120
  _ _ _ _ _ 3 =>5!=5*4*3*2*1=120
  _ _ _ _ _ 4 =>2*4*3*2*1=48 (Tenth digit can have either 2 or 6)
  _ _ _ _ _ 5 =>5!=5*4*3*2*1=120
  _ _ _ _ _ 6 =>5!=5*4*3*2*1=120
  Total: 120*5 + 48 =648 ans
  
• 10 years agoHelpfull: Yes(1) No(2)