how many six digits numbers can be formed using the digits 1to 6, without repetition. such that the number is divisible by the digits at its unit place

• _ _ _ _ _ 1 =>5*4*3*2*1
  _ _ _ _ _ 2 =>5*4*3*2*1
  _ _ _ _ _ 3 =>5*4*3*2*1
  _ _ _ _ _ 4 =>2*4*3*2*1=48 (Tens place can have either 2 or 6)
  _ _ _ _ _ 5 =>5*4*3*2*1
  _ _ _ _ _ 6 =>5*4*3*2*1
  Total: 120*5 + 48 =648 ans
  
• 10 years agoHelpfull: Yes(32) No(6)
• divisible by 1 --120
  2---120
  3---120
  4----48
  5----120
  6 ---120
  total = 648ans.
• 10 years agoHelpfull: Yes(8) No(2)
• unit digit 1:120
  2:120
  3:120
  4:48
  5:120
  6:120
  total=648
• 10 years agoHelpfull: Yes(3) No(1)
• 2*4*3*2*1=48 (Tens place can have either 2 or 6)
  
  i cannot understand this. why tenth digit used in this.
• 10 years agoHelpfull: Yes(1) No(0)
• plz explain step by step....cant understand
  
• 10 years agoHelpfull: Yes(0) No(1)
• numbers ending with 1,2,5 and 6 will respectively be divisible by their unit digit. hence total no. ending with 1,2,5 and 6 are 4*(5!)=480.
  and numbers ending with 3 are not all divisible by 3.so consider the series 3,33,63,93------.here we get the clue that no. ending with 3 and divisible by 3 are located at a distance of 30.
  now total no. divisible by 3=120 and out of them divisible by 3 are 120/30=4.
  similarly no. ending with 4 and divisible by 4 are located at a distance of 20.
  now total no. divisible by 4=120 and out of them divisible by 4 are 120/20=6.
  hence total numbers divisible by by the digits at unit place are =480+6+4=490.
  hence 490 ans.
• 10 years agoHelpfull: Yes(0) No(1)
• -----1=120
  -----2=120
  -----5=120
  -----3=24 (only numbers ending with ----63)
  -----6=24 (only numbers ending with ----36)
  -----4=48 (only numbers ending with ----24 and ----64)
  total = 120+120+120+24+24+48 = 456 ans
• 10 years agoHelpfull: Yes(0) No(1)